Question 11 Mark
The value of $\Big(\frac{1}{2^3}\Big)^3$ is equal to _________.
AnswerThe value of $\Big(\frac{1}{2^3}\Big)^2$ is equal to $\frac{1}{2^6}$.
Solution:
Using law of exponents, $(am)n = (a)mn [$$\because$ a is non-zero integer$]$
$\because$ $\Big(\frac{1}{2^3}\Big)^2=\Big(\frac{1^3}{2^3}\Big)^2=\Big(\frac{1}{2}\Big)^{3\times2}$
$=\Big(\frac{1}{2}\Big)^6=\frac{1}{2^6}$ [$\because$ $(1)^m= 1]$
Hence,
$\Big(\frac{1}{2^3}\Big)^2$ = $\frac{1}{2^6}$
View full question & answer→Question 21 Mark
The value of $3 \times 10^{-7}$ is equal to ________.
AnswerThe value of $3 \times 10^{-7}$ is equal to $0.0000003.$
Solution:
Given,
$3 \times 10^{-7}= 3.0 × 10^{-7}$
Now, placing decimal seven place towards left of original position, we get $0.0000003.$
Hence,
The value of $3 \times 10^{-7}$ is equal to $0.0000003.$
View full question & answer→Question 31 Mark
The usual form for $2 × 10^{-2}$ is not equal to $0.02.$
AnswerFalse.Solution:
For usual form, $2\times10^{-2}=2\times\frac{1}{10^2}$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}\Big]$ $=\frac{2}{100}=0.02$
View full question & answer→Question 41 Mark
To add the numbers given in standard form, we first convert them into numbers with ________ exponents.
AnswerTo add the numbers given in standard form, we first convert them into numbers with Equal exponents.
Solution:
To add the numbers given in standard form, we first convert them into numbers with equal exponents.
$2.46 \times 10^6+24.6 \times 105 $
$ =2.46 \times 10^5+2.46 \times 10^6 $
$ =4.92 \times 10^6 $
View full question & answer→Question 51 Mark
$\left[2^{-1}+3^{-1}+4^{-1}\right]^0=$________.
Answer$\left[2^{-1}+3^{-1}+4^{-1}\right]^0=1$.
Solution:
Using law of exponents, $a^0= 1$ [$\because$ a is non-zero integer]
$\therefore$ $\left[2^{-1}+3^{-1}+4^{-1}\right]^0=1$
Hence,
$\left[2^{-1}+3^{-1}+4^{-1}\right]^0=1$
View full question & answer→Question 61 Mark
$\Big(\frac{1}{2}\Big)^{-2}\div\Big(\frac{1}{2}\Big)^{-3}$
Answer$\Big(\frac{1}{2}\Big)^{-2}\div\Big(\frac{1}{2}\Big)^{-3}$
$=\Big(\frac{1}{2}\Big)^{-2+3}$ $=\frac{1}{2}$
$\left[\because a^m \div a^n=(a)^{m-n}\right]$
View full question & answer→Question 71 Mark
On multiplying _________ by $2^{-5}$ we get $2^5$.
AnswerOn multiplying $2^{10}$ by $2^{-5}$ we get $2^5$.
Solution:
Let $x$ be multiplied by $2^{-5}$ we get $2^5$.
So, $x × 2^{-5}= 2^5$
$\Rightarrow\text{x}\times\frac{1}{2^5}=2^5\ \Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{-\text{m}}}\Big]$ $\therefore$ $\text{x}=2^5\times2^5=(2)^{5+5}=2^{10}$ [$\because$ $a^m× a^n= (a)^{m+n}$]
View full question & answer→Question 81 Mark
Find two repeater machines that will do the same work as $a(x81)$ machine.
AnswerTwo repeater machines that do the same work as $(x81)$ are $(x3^4)$ and $(x9^2)$. Since, factor of $81$ are.$3$ and $9.$
View full question & answer→Question 91 Mark
Simplify: $\frac{49\times\text{z}^{-3}}{7^{-3}\times10\times\text{z}^{-5}}(\text{z}\neq0)$
Answer$\frac{49\times\text{z}^{-3}}{7^{-3}\times10\times\text{z}^{-5}}$
$=\frac{(7)^2\times\text{z}^{-3}}{7^{-3}\times10\times\text{z}^{-5}}$
$=\frac{(7)^{2+3}\times\text{z}^{-3+5}}{10}$ $\left[\because a^m \div a^n=(a)^{m-n}\right]$
$=\frac{(7)^5\text{z}^2}{10}=\frac{7^5}{10}\text{z}^2$
View full question & answer→Question 101 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $\left(a^m \times a^n=a^{m+n}\right)$$[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $\big(\frac{1}{2}\big)^2\times\big(\frac{1}{3}\big)^2$
$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{3}\times\frac{1}{3}=\frac{1}{2\times3\times2\times3}=\frac{1}{(6)^2}$
so, $\Big(\text{x}\big(\frac{1}{6}\big)^2\Big)$ single machine can do the same work.
View full question & answer→Question 111 Mark
$5^0$= ___________.
Answer$5^0= 1.$
Solution:
Using law of exponents, $a^0= 1$ [$\because$ a is non-zero integer] $\therefore$ $5^0= 1$ Hence, $5^0= 1$
View full question & answer→Question 121 Mark
Very small numbers can be expressed in standard form by using _________ exponents.
AnswerVery small numbers can be expressed in standard form by using negative exponents.
Solution:
Very small numbers can be expressed in standard form by using negative exponents.
i.e. $0.000023=2.3 \times 10^{-3}$
View full question & answer→Question 131 Mark
Neha needs to stretch some sticks to $25^2$ times their original lengths, but her $(×25)$ machine is broken. Find a hook-up of two repeater machines that will do the same work as $a(×25^2)$ machine. To get started, think about the hookup you could use to replace the $(×25) $ machine.
AnswerWork done by single machine $(x25^2) = 25 \times 25 = 625$ or $5 \times 5 \times 5 \times 5$ or $52 \times 52.$
Hence, $(x5^2)$ and $(x5^2)$ hook-up machine can replace the $(x25)$ machine.
View full question & answer→Question 141 Mark
Use the properties of exponents to verify that each statement is true. $4^{\text{n}-1}=\frac{1}{4}(4)^{\text{n}}$
Answer$4^{\text{n}-1}=\frac{1}{4}(4)^{\text{n}}$ $\text{LHS} = 4^{\text{n} - 1} = 4^\text{n} + 4^1$ $[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=(\text{a})^{\text{m}-\text{n}}]$ $=\frac{4^{\text{n}}}{4}=\text{RHS}$
View full question & answer→Question 151 Mark
A half-life is the amount of time that it takes for a radioactive substance to decay to one half of its original quantity. Suppose radioactive decay causes $300$ grams of a substance to decrease to $300 × 2^{-3}$ grams after $3$ half-lives. Evaluate $300 × 2^{-3}$ to determine how many grams of the substance are left.
AnswerSince, $300g$ of a substance is decrease to $300 × 2^{-3}g$ after $3$ half-lives.
So, we have to evalute $300\times2^{-3}=\frac{300}{8}=37.5\text{g}$ $[\because2^3=8]$
View full question & answer→Question 161 Mark
Write $0.000005678$ in the standard form.
AnswerFor standard form, $0.000005678$
$ =0.5678 \times 10^{-5} $
$ =5.678 \times 10^{-5} \times 10^{-1} $
$ =5.678 \times 10^{-6} $
Hence, $ =5.678 \times 10^{-6} $ is the standard form of $0.000005678.$
View full question & answer→Question 171 Mark
$a^p× b^q= (ab)^{pq}$
Answer$RHS = (ab)^{pq}$
Using law of exponents, $(ab)^m= a^m× b^m$ [$\because$ a is non-zero integer]
$\therefore$ $(ab)^{pq}= (a)^{pq}× (b)^{pq}$
$LHS ≠ RHS$
View full question & answer→Question 181 Mark
The expression for $3^5$ with a negative exponent is _________.
AnswerThe expression for $3^5$ with a negative exponent is $\frac{1}{3}^{-5}$.
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$[$\because$ a is non-zero integer]
$\therefore$ $3^5=\frac{1}{3}^{-5}$
Hence,
The expression for $3^5$ with a negative exponents is $\frac{1}{3}^{-5}$
View full question & answer→Question 191 Mark
The value of $\left[2^{-1} \times 3^{-1}\right]^{-1}$ is _________.
AnswerThe value of $\left[2^{-1} \times 3^{-1}\right]^{-1}$ is 6.
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$ [$\because$ a is non-zero integer]
$\therefore$ $\big[2^{-1}\times3^{-1}\big]^{-1}=\Big(\frac{1}{2}\times\frac{1}{3}\Big)^{-1}$
$=\Big(\frac{1}{6}\Big)^{-1}=6$
Hence,
$\big[2^{-1}\times3^{-1}\big]^{-1}=6$
View full question & answer→Question 201 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $\left(a^m \times a^n=a^{m+n}\right)$$[\therefore$ a is non-zero integer$]$Repeator machine can do the work is equal to $3^y \times 3^y=3^{2 y}$.
so, $(x3^{2y})$ single machine can do the same work.
View full question & answer→Question 211 Mark
For hook-up, determine whether there is a single repeater machine that will do the same work. If so, describe or draw it.
AnswerUsing law of exponents, $\left(a^m \times a^n=a^{m+n}\right)$$[\because$ a is non-zero integer$]$
Hook-up machine work $=2^2\times\big(\frac{1}{3}\big)^3\times5^4=4\times\frac{1}{27}\times625 $$=\frac{2500}{27}=92.59$
So, it is not possible for a single machine can do the same work.
View full question & answer→Question 221 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $\left(a^m \times a^n=a^{m+n}\right)$$[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $2^2× 2^3× 2^4= 2^9$
so, $(x2^9)$ single machine can do the same work.
View full question & answer→Question 231 Mark
Supply the missing information for diagram.
AnswerIf $5\ cm$ long piece is inserted in single machine, then it produce same $15\ cm$ long piece. So, it is $(x5)$ repeated machine. $\because \ ?=5$
View full question & answer→Question 241 Mark
$24.58 = 2 \times 10 + 4 \times 1 + 5 \times 10 + 8 \times 100$
Answer$R H S = 2 \times 10 + 4 \times 1+ 5 \times 10 + 8 \times 100$
$= 20 + 4 + 50 + 800 = 874$
$L H S \neq R H S$
View full question & answer→Question 251 Mark
The value of $\frac{1}{4^{-2}}$ is equal to $16.$
AnswerUsing law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$ [$\because$ a is non-zero integer] $\therefore \frac{1}{4^{-2}}=4^2$
$=4\times4=16$
View full question & answer→Question 261 Mark
Simplify: $\Big[\big(\frac{4}{3}\big)^{-2}-\big(\frac{3}{4}\big)^2\Big]^{(-2)}$
Answer$\Big[\big(\frac{4}{3}\big)^{-2}-\big(\frac{3}{4}\big)^2\Big]^{(-2)}$$=\Big[\big(\frac{3}{4}\big)^{2}-\big(\frac{3}{4}\big)^2\Big]^{(-2)}=[0]^{-2}=0$
View full question & answer→Question 271 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $\left(a^m \times a^n=a^{m+n}\right)$$[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $7^{10}× 7^{50}×7^{1}= 7^{61}$.
so, $(x7^{61})$ single machine can do the same work.
View full question & answer→Question 281 Mark
The multiplicative inverse of $\Big(\frac{3}{2}\Big)^2$ is not equal to $\Big(\frac{2}{3}\Big)^{-2}$.
AnswerTrue. Solution:
a is called the multiplicative inverse of $b,$ if $a \times b = 1.$
put $b =\Big(\frac{3}{2}\Big)^2$
So, $\text{a}\times\Big(\frac{3}{2}\Big)^2=1$
$\Rightarrow\text{a}=\Big(\frac{3}{2}\Big)^{-2}$
View full question & answer→Question 291 Mark
The value of $\left[1^{-2}+2^{-2}+3^{-2}\right] \times 6^2$ is ________ .
AnswerThe value of $\left[1^{-2}+2^{-2}+3^{-2}\right] \times 6^2$ is $49.$
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer]
$\therefore \big[1^{-2}+2^{-2}+3^{-2}\big]\times6^2$
$=\Big[\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^3}\Big]\times6^2$
$=\Big[1+\frac{1}{4}+\frac{1}{9}\Big]\times6^2$
$=\Big(\frac{36+9+4}{36}\Big)\times6^2$
$=\Big(\frac{49}{36}\Big)\times6^2=\Big(\frac{7}{6}\Big)^2\times6^2$
$(7)^2\times6^{-2}\times6^2$
$=(7)^2\times6^{2-2}=(7)^2\times6^0=49$
$[a^0= 1]$
Hence,$ [1^{-2}+ 2^{-2}+ 3^{-2}] \times 6^2= 49$
View full question & answer→Question 301 Mark
If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use $(x1)$ machines.
Answer$x1111 = 101 \times 11$ hook-up machine.
View full question & answer→Question 311 Mark
If $a = -1, b = 2,$ then find the value of the following:
$a^b× b^2$
AnswerGiven, $a^b× b^2$
If $a = -1$ and $b = 2,$ then $(-1)^2 \times(-2)^{-1}=1\times\frac{1}{2}=\frac{1}{2}$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 321 Mark
The standard form of $12340000$ is ______.
AnswerThe standard form of $12340000$ is $1.234 × 10^7$.
Solution:
For standard form, $12340000 = 1234 × 10^4$
$= 1234 × 10^4× 10^3$
$= 1234 × 10^7$
Hence, The standard form of $12340000$ is $1.234 × 10^7$.
View full question & answer→Question 331 Mark
The value of $5^{-2}$ is equal to $25.$
AnswerFalse.Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer] $\therefore$ $5^{-2}=\frac{1}{5^2}=\frac{1}{25}$
View full question & answer→Question 341 Mark
If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use $(x1)$ machines.
AnswerSingle machine work $= 100$
Hook-up machine of prime base number that do the same work down by $x100$
$= 2^2× 5^2$
$= 4 × 25$
$= 100$
View full question & answer→Question 351 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $(a^m× a^n= a^{m+n})$ $[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $100^2× 100^{10}$.
so, $(x100^{12})$ single machine can do the same work.
View full question & answer→Question 361 Mark
The standard form for $32,50,00,00,000$ is __________.
AnswerThe standard form for $32,50,00,00,000$ is $3.25 × 10^{10}$
Solution:
For standard form, $32500000000 = 3250 × 10^2× 10^2× 10^3= 3250 × 10^7= 3.250 × 10^{10}$ or $3.25 × 10^{10}$
Hence, The standard form for $32500000000$ is $3.25 × 10^{10}$.
View full question & answer→Question 371 Mark
An inch is approximately equal to $0.02543$ metres. Write this distance in standard form.
AnswerStandard form of $0.02543m = 0.2543\ x\ 10^{-1}m = 2.543\ x\ 10^{-2}m$.
Hence, standard form of $0.025434s$ $2.543\ x\ 10^{-2}$ m.
View full question & answer→Question 381 Mark
$10^{-2}=\frac{1}{100}$
AnswerTrue.Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{}\text{a}^\text{m}}$ [$\because$ a is non-zero integer] $\therefore$ $10^{-2}=\frac{1}{10^2}$ $[\because10^2=10\times10]$ $=\frac{1}{10\times10}=\frac{1}{100}$
View full question & answer→Question 391 Mark
The value of $\left[3^{-1} \times 4^{-1}\right]^2$ is _________.
AnswerThe value of $\left[3^{-1} \times 4^{-1}\right]^2$ Is $\frac{1}{144}$.
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer]
$\therefore$ $\big[3^{-1}\times4^{-1}\big]^2=\Big(\frac{1}{3}\times\frac{1}{4}\Big)^2=\Big(\frac{1}{12}\Big)^2$
$=12^{-2}=\frac{1}{144}$
Hence,
$\big[3^{-1}\times4^{-1}\big]^2=\frac{1}{144}$
View full question & answer→Question 401 Mark
$100^{-10}$
Answer$100^{-10}$
$=\frac{1}{100^{10}}$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}\Big]$
View full question & answer→Question 411 Mark
The usual form for $2.3 \times 10^{-10}$ is ____________.
AnswerThe usual form for $2.3 \times 10^{-10}$ is $0.00000000023.$
Solution:
For usual form, $2.3 \times 10^{-10}=0.23 \times 10^{-11}=0.00000000023$
Hence,
The usual form for $2.3 \times 10^{-10}$ is $0.00000000023.$
View full question & answer→Question 421 Mark
If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use $(x1)$ machines.
Answer$x99 = 3^2× 111$ hook-up machine.
View full question & answer→Question 431 Mark
If $a = -1, b = 2,$ then find the value of the following:
$a^b- b^a$
AnswerGiven, $a^b- b^a$
If $a = -1$ and $b = 2,$ then $(-1)^2 + (2)^{-1}=1-\frac{1}{2^1}$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{n}}}\Big]$
$=\frac{2-1}{2}=\frac{1}{2}$
View full question & answer→Question 441 Mark
If $a = -1, b = 2,$ then find the value of the following:
$a^b+ b^a$
AnswerGiven, $a^b+ b^a$
If $a = -1$ and $b = 2,$ then $(-1)^2 + (-2)^{-1}=1+\frac{1}{2}$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
$=\frac{2+1}{2}=\frac{3}{2}$
View full question & answer→Question 451 Mark
Shikha has an order from a golf course designer to put palm trees through a $(x23)$ machine and then through a $(x33)$ machine. Shethinks she can do the job with a single repeater machine. What single repeater machine should she use$?$ 
AnswerThe work done by hook-up machine is equal to $2 × 2 × 2 × 3 × 3 × 3 = 216 = 6^3$ So, she should use $(x6^3)$ single machine for the purpose.
View full question & answer→Question 461 Mark
Express $16^{-2}$ as a power with the base $2.$
Answer$\because$ $2 × 2 × 2 × 2 = 16 = 2^4$
$\because$ $16^{-2}= (2^4)^{-2}= (2)^{4\times {(-2)}}$
[$\because$ $(a^m)^n= (a)^{mn}] = (2)^{-8}$
View full question & answer→Question 471 Mark
$(-5)^{-2} \times(-5)^{-3}=(-5)^{-6}$
AnswerFalse.
Solution:
$LHS =(-5)^{-2} \times(-5)^{-3}$
Using law of exponents, $a^m \times a^n=(a)^{m+n}$ [$\because$ a is non-zero integer]
$\therefore(-5)^{-2} \times(-5)^{-3}=(-5)^{-2-3}=(-5)^{-5}$
$LHS ≠ RHS$
View full question & answer→Question 481 Mark
The usual form of $3.41 × 10^6$ is ___________.
AnswerThe usual form of $3.41 × 10^6$ is $3410000.$
Solution:
For usual form, $3.41 × 10^6= 3.41 × 10 × 10 × 10 × 10 × 10 × 10 = 3410000$
Hence, The usual form of $3.41 × 10^6$ is $3410000.$
View full question & answer→Question 491 Mark
If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use (x1) machines.
Answer$x37$ machine cannot do the same work.
View full question & answer→Question 501 Mark
$2^{-2} \times 2^{-3}$
Answer$2^{-2} \times 2^{-3}$
$=(2)^{-2-3}\left[\because a^m \times a^n=(a)^{m+n}\right] $
$=(2)^{-5}$
View full question & answer→Question 511 Mark
The standard form for 203000 is $2.03 × 10^5$.
AnswerTrue.
Solution:
For standard form, $203000 = 203 × 10 × 10 × 10 = 203 × 10^3$
$= 2.03 × 10^2× 10^3$
$= 2.03 × 10^5$
View full question & answer→Question 521 Mark
What will the following machine do to a $2\ cm$ long piece of chalk?
AnswerThe machine produce $x1^{100}= 1$
So, if we insert $2\ cm$ long piece of chalk in that machine, the piece of chalk remains same.
View full question & answer→Question 531 Mark
$\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=$ _________.
Answer$\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=\Big(\frac{2}{13}\Big)^{-36}$
Solution:
Using laws of exponents, $a^m \div a^n=(a)^{m-n}$ and $a^m \times a^n=(a)^{m+n}$[$\therefore$ a is non-zero integer]
$\therefore$ $\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}$
$=\bigg[\Big(\frac{2}{13}\Big)^{-6-3}\bigg]^3\times\Big(\frac{2}{13}\Big)^{-9}$
$=\Big(\frac{2}{13}\Big)^{-27}\times\Big(\frac{2}{13}\Big)^{-9}$
$=\Big(\frac{2}{13}\Big)^{-27-9}=\Big(\frac{2}{13}\Big)^{-36}$
Hence, $\bigg[\Big(\frac{2}{13}\Big)^{-6}\div\Big(\frac{2}{13}\Big)^{3}\bigg]\times\Big(\frac{2}{13}\Big)^{-9}=\Big(\frac{2}{13}\Big)^{-36}$
View full question & answer→Question 541 Mark
The expontential form for $(-2)^4\times\Big(\frac{5}{2}\Big)^4$ is $5^4$.
AnswerUsing law of exponents, $a^m \div a^n=(a)^{m-n}$ and $\Big(\frac{\text{a}}{\text{b}}\Big)^\text{m}=\frac{\text{a}^\text{m}}{\text{b}^\text{m}}$
[$\because a$ and $b$ are non-integers]
$\therefore$ $(-2)^4\times\Big(\frac{5}{2}\Big)^4=(2)^4\times\frac{(5)^4}{(2)^4}$
$=(2)^{4-4}\times5^4$
$=2^0\times5^4=5^4$ [$\because$ $(-a^m) = (a^m)$, if $m$ is an even number]
View full question & answer→Question 551 Mark
By multiplying $\Big(\frac{5}{3}\Big)^4$ by ________ we get $5^4$.
AnswerBy multiplying $\Big(\frac{5}{3}\Big)^4$ by ________ we get $5^4$.
Solution:
Let $x$ be multiplied with $\Big(\frac{5}{3}\Big)^4$ to get $5^4$.
So,
$\Big(\frac{5}{3}\Big)^4\times\text{x}=5^4$
$\therefore$ $\text{x}=\frac{5^4}{\Big(\frac{5}{3}\Big)^4}$
$=5^4\times3^4\times5^{-4}=(5)^{4-4}\times3^4$ $\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}},\text{a}^\text{m}\times\text{a}^{\text{n}}=(a)^{\text{m}+\text{n}}\Big]$
$=5^o\times3^4=1\times81=81$ [$\because$ $a^0= 1$]
View full question & answer→Question 561 Mark
$\Big(-\frac{8}{2}\Big)^0=0$
Answer$LHS =\Big(-\frac{8}{2}\Big)^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore$ $\Big(-\frac{8}{2}\Big)=1$
$LHS ≠ RHS$
View full question & answer→Question 571 Mark
The value for $(-7)^6 \div 7^6$ is _________.
Answer Using law of exponents, $a^m \div a^n=(a)^{m-n}[\because$ a is non-zero integer$]$
$\therefore(-7)^6 \div 7^6=(-7)^6 \div 7^6\left[\left(-a^m\right)=\left(a^m\right)\right.$, if $m$ is an even number$]$
$=(7)^{6-6}=(7)^0=1\left[\because a^0=1\right]$
Hence,
$(-7)^6 \div 7^6=1$
View full question & answer→Question 581 Mark
The expression for $4^{-3}$ as a power with the base $2$ is $2^6$.
AnswerFalse.Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ $\therefore$ $4^{-3}=\frac{1}{4^3}$ $=\frac{1}{(2^2)^3}=\frac{1}{(2)^6}$ $[\because2\times2=4,(\text{a}^\text{m})^\text{n}=(\text{a})^\text{mn}]$
View full question & answer→Question 591 Mark
$\Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{2}{3}\Big)^{10}$
AnswerFalse.Solution:
$LHS \Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}$
Using law of exponents, $a^m \times a^n=(a)^{m+n}[\because$ a is non-zero integer$]$
$\therefore$ $\Big(\frac{2}{3}\Big)^{-2}\times\Big(\frac{2}{3}\Big)^{-5}=\Big(\frac{2}{3}\Big)^{-2-5}$
$=\Big(\frac{2}{3}\Big)^{-7}$
$LHS ≠ RHS$
View full question & answer→Question 601 Mark
Express $3^{-5}\times 3^{-4}$ as a power of $3$ with positive exponent.
AnswerUsing laws of exponents, $a^m \times a^n=(a)^{m+n}$ and $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}[\because$ a is not-zero integer$]$
$\therefore$ $3^{-5}\times3^{-4}=(3)^{-5-4}$
$=(3)^{-9}=\frac{1}{3^9}$
View full question & answer→Question 611 Mark
Supply the missing information for diagram.
AnswerIf $x\ cm$ long piece is inserted in $(x4)$ and $(x3)$ hooked machine, repeated machine, then it will produce $36\ cm$ long piece. So, $x \times 4 \times 3 = 36$
$ \Rightarrow x = 3\ cm$
View full question & answer→Question 621 Mark
Find three machines that can be replaced with hook-ups of $(x^5)$ machines.
AnswerSince, $5^2= 25, 5^3= 125, 5^4= 625$
Hence, $(x5^2), (x5^3)$ and $(x5^4)$ machine can replace $(x^5)$ hook-up machine.
View full question & answer→Question 631 Mark
By solving $(6^0- 7^0) × (6^0+ 7^0)$ we get ________.
AnswerUsing law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore$ $(6^0- 7^0) × (6^0+ 7^0) = (1 - 1) × (1 + 1)$
$= 0 × 2 = 0$
Hence,
$(6^0- 7^0) × (6^0+ 7^0) = 0$
View full question & answer→Question 641 Mark
If $a = -1, b = 2, $ then find the value of the following: $a^b÷ b^a$
AnswerGiven, $a^b+ b^a$
If $a = -1$ and $b = 2,$ then $(-1)^2 \div(-2)^{-1}=1+\frac{1}{2^1}={1}\times{2}=2$
$\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 651 Mark
Use the properties of exponents to verify that each statement is true. $\frac{1}{4}(2^{\text{n}})=2^{\text{n}-2}$
Answer$\frac{1}{4}(2^{\text{n}})=2^{\text{n}-2}$ $\text{RHS} = 2^{\text{n} - 2} = 2^\text{n} + 2^2 $ $[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=(\text{a})^{\text{m}-\text{n}}]$ $=\frac{2^{\text{n}}}{4}=\text{LHS}$
View full question & answer→Question 661 Mark
Supply the missing information for diagram.
AnswerIf $5\ cm$ long piece is inserted in single machine, then it produce same $5\ cm$ long piece. So, it is $(x1)$ repeated machine. $\because \ ?=1$
View full question & answer→Question 671 Mark
The standard form of $\Big(\frac{1}{100000000}\Big)$ is ______.
AnswerThe standard form of $\Big(\frac{1}{100000000}\Big)$ is $1.0 \times 10^{-8}$
Solution:
For standard form, $\Big(\frac{1}{100000000}\Big)=\frac{1}{1\times10^8}=\frac{1}{10^8}$
$=1 \times 10^{-8}=1.0 \times 10^{-8}$
Hence, Standard form of $\Big(\frac{1}{100000000}\Big)$ is $1.0 \times 10^{-8}$.
View full question & answer→Question 681 Mark
Use the properties of exponents to verify that each statement is true. $25(5^{\text{n}-2})=5^{\text{n}}$
Answer$25(5^{\text{n}-2})=5^{\text{n}}$ $\text{LHS}=25(5^{\text{n}-2})=5^{\text{2}}(5^{\text{n}}+5^2)$ $=5^2\times5^\text{n}\times\frac{1}{5^2}=5^{\text{n}}=\text{RHS}$
View full question & answer→Question 691 Mark
By multiplying $(10)^5$ by $(10)^{-10}$ we get ________.
AnswerBy multiplying $(10)^5$ by $(10)^{-10}$ we get $(10)^{-5}$.
Solution:
Using law of exponents, $(a)^m× (a)^n= (a)^{m+n}$ [$\because$ a is non-zero integer]
Similarly,
$(10)^5× (10)^{-10}+ (10)^{5-10}$
$= (10)^{-5}$
Hence, $(10)^5× (10)^{-10}=(10)^{-5}$
View full question & answer→Question 701 Mark
$(-4)^{-4}× (4)^{-1}= (4)^5$
Answer$LHS = (-4)^{-4}× (4)^{-1}$
Using law of exponents, $a^{m}× a^n= (a)^{m+n}[\because$ a is non-zero integer$]$
$\therefore$ $(-4)^{-4}× (4)^{-1}= (4)^{-4}× (4)^{-1}$
$= (-4)^{-4-1}$
$= (-4)^{-5}$
$LHS ≠ RHS$
View full question & answer→Question 711 Mark
$5^5× 5^{-5}$= __________.
Answer$5^5× 5^{-5}= 1.$
Solution:
Using law of exponents, $a^m× a^n= (a)^{m+n}$[$\because$ a is non-zero integer]
$\therefore$ $5^5× 5^{-5}= (5)^{5-5}$
$= (5)^0= 1$ [$\therefore$ $a^0= 1$]
Hence,
$5^5× 5^{-5}= 1$
View full question & answer→Question 721 Mark
The standard form for $0.000037$ is $3.7 × 10^{-5}$.
AnswerFor standard form,$ 0.000037 = 0.37 × 10^{-4}$
$= 3.7 × 10^{-5}$
View full question & answer→Question 731 Mark
The volume of the Earth is approximately $7.67 × 10^{-7}$ times the volume of the Sun. Express this figure in usual form.
AnswerGiven, volume of the Earth is $7.67 × 10^{-7}$ times the volume of the sun Usual form of $7.67 × 10^{-7}= 0.000000767.$ $[\because$ placing decimal $7$ places towards the left of original position$]$
View full question & answer→Question 741 Mark
$5^0= 5$
Answer$LHS = 5^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore 5^0= 1$
$LHS ≠ RHS$
View full question & answer→Question 751 Mark
$a^3 \times a^{-10}=$= __________.
Answer$a^3 \times a^{-10}= a^{-7}.$
Solution:
Given,
Using law of exponents, $a^m \times a^n=(a)^{m+n}$[$\because$ a is non-zero integer]
Similarly,
$a^3 \times a^{-10}=(a)^{3-10}$
$=(a)^{-7}$
Hence,
$a^3 \times a^{-10}=a^{-7}$
View full question & answer→Question 761 Mark
$(-7)^{-4} \times(-7)^2=(-7)^{-2}$
Answer$LHS =(-7)^{-4} \times(-7)^2$
Using law of exponents, $a^m \times a^n=(a)^{m+n}$ [$\because$ a is non-zero integer]
$\therefore(-7)^{-4} \times(-7)^2=(-7)^{-4+2}$
$=(-7)^{-2}$
$LHS ≠ RHS$
View full question & answer→Question 771 Mark
$(-6)^0= -1$
Answer$LHS = (-6)^0$
Using law of exponents, $a^0= 1 [ \because$ a is non-zero integer$]$
$\therefore (-6)^0= -1$
$LHS ≠ RHS$
View full question & answer→Question 781 Mark
The usual form of $2.39461 × 10^6$ is ________.
AnswerThe usual form of $2.39461 \times 10^6$ is $2394610.$
Solution:
For usual form, $2.39461 \times 10^6= 2394610 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$
$= 2394610$
Hence,
The usual form of $2.39461 \times 10^6$ is $2394610.$
View full question & answer→Question 791 Mark
On dividing $8^5$ by _________ we get $8.$
AnswerLet $8^5$ be divided by $x$ to get $8.$
So, $8^5÷ x = 8$
$\Rightarrow8^5\times\frac{1}{\text{x}}=8$
$\Rightarrow\frac{8^5}{8}=\text{x}$
$\therefore$ $\text{x}=\frac{8^5}{8^1}=8^{5-1}=8^4[\because$ $a^m÷ a^n= (a)^{m-n}]$
View full question & answer→Question 801 Mark
$329.25=3 \times 10^2+2 \times 10^1+9 \times 10^0+2 \times 10^{-1}+5 \times 10^{-2}$
Answer$RHS =3 \times 10^2+2 \times 10^1+9 \times 10^0+2 \times 10^{-1}+5 \times 10^{-2}$
$=3\times10\times10+2\times10+9\times1+\frac{2}{10}+\frac{5}{10\times10}[\because a^0= 1] $
$=300+20+9+0.2-0.05=329.25$
$\therefore LHS = RHS$
View full question & answer→Question 811 Mark
Find a single repeater machine that will do the same work as hook-up.
AnswerUsing law of exponents, $ (a^m× a^n= a^{m+n})$ $[\therefore$ a is non-zero integer$]$
Repeator machine can do the work is equal to $2^2\times\big(\frac{1}{2}\big)^3\times2^4=2^6\times\frac{1}{2^3}=2^3$$\Big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{\text{n}}}=\text{a}^{\text{m}-\text{n}}\Big]$
so, $(x2^3)$ single machine can do the same work.
View full question & answer→Question 821 Mark
Very large numbers can be expressed in standard form by using _________ exponents.
AnswerVery large numbers can be expressed in standard form by using positive exponents.
Solution:
Very large numbers can be expressed in standard form by using positive exponents,
$ 23000=23 \times 10^3$
$ =2.3 \times 10^3 \times 10^1 $
$=2.3 \times 10^4 $
View full question & answer→Question 831 Mark
The standard form for $0.000000008$ is __________.
AnswerThe standard form for $0.000000008$ is $8.0 × 10^{-9}$.
Soluiton:
For standard form, $0.000000008 = 0.8 \times 10^{-8}= 8 \times 10^{-9}= 8.0 \times 10^{-9}$
Hence,
The standard form for $0.000000008$ is $8.0\times 10^{-9}$
View full question & answer→Question 841 Mark
Find the value $ \left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
AnswerUsing law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}[ \because$ a is non-zero integer$]$
$\therefore\left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
$=\Big(\frac{1}{4}+\frac{1}{3}+\frac{1}{36}\Big)^{-1}$
$=\Big(\frac{9+12+1}{36}\Big)^{-1}[\because LCM$ of $4, 3,$ and $36 = 36]$
$=\Big(\frac{22}{36}\Big)^{-1}=\frac{36}{22}=\frac{18}{11}$
Hence,
$\Big[4^{-1}+3^{-1}+6^{-2}\Big]^{-1}=\frac{18}{11}$
View full question & answer→Question 851 Mark
$a^m× b^m= (ab)^m$
Answer$LHS = a^m× b^m= (a × b)^m= (ab)^m$
$[$by law of exponents$]$
View full question & answer→Question 861 Mark
Supply the missing information for diagram.
AnswerIf $1.25\ cm$ long piece is inserted in $(x4)$ repeated machine, then it will produce $1.25 × 4 = 10\ cm$ long piece. $\because \ ?=10\text{cm}$
View full question & answer→Question 871 Mark
Large numbers can be expressed in the standard form by using positive exponents.
Answere.g. $2360000 = 236 \times 10 \times 10 \times 10 \times 10$
$ =236 \times 10^4 $
$ =2.36 \times 10^4 \times 10^2 $
$ =2.36 \times 10^6 $
View full question & answer→Question 881 Mark
The multiplicative inverse of $(-4)^{-2}$ is $(4)^{-2}$.
Answer$a$ is called the multiplicative inverse of b, if $a \times b = 1.$
Put $b = (-4)^{-2}$
$\therefore$ $a × (-4)^{-2}= 1$
$\Rightarrow\text{a}=\frac{1}{(-4)^{-2}}=(-4)^2$
View full question & answer→Question 891 Mark
$\frac{\text{x}^\text{m}}{\text{y}^\text{m}}=\Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}$
Answer$RHS =\Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}$
Using law of exponents, $\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}}=\frac{\text{a}^\text{m}}{\text{b}^\text{m}}$ and $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}} [\because$ a is non-zero integers$]$
$\therefore \Big(\frac{\text{y}}{\text{x}}\Big)^{-\text{m}}=\frac{\text{y}^{-\text{m}}}{\text{y}^{-\text{m}}}$
$=\frac{\text{x}^\text{m}}{\text{y}^\text{m}} $
$LHS = RHS$
View full question & answer→Question 901 Mark
$ 3^5 \div 3^{-6}$ can be simplified as __________.
Answer$3^5 \div 3^{-6}$ can be simplified as $(3)^{11}$.
Solution:
Given,
$3^5 \div 3^{-6}=(3)^{5-(-6)}$
$=(3)^{5+6}=(3)^{11}\left[\because a^m+a^n=(a)^{m-n}\right]$
Hence,
$3^5 \div 3^{-6}$ can be simplified as $(3)^{11}$.
View full question & answer→Question 911 Mark
The expression for $8^{-2}$ as a power with the base $2$ is _________.
AnswerGiven,
$8^{-2}$, Where we can write $8 = 2 \times 2 \times 2$
$\therefore(2 \times 2 \times 2)^{-2}=(2)^{3 \times(-2)}=(2)^{-6}$
Hence,
$8^{-2}=(2)^{-6}$
View full question & answer→Question 921 Mark
$(-2)^0= 2$
Answer$LHS = (-2)^0$
Using law of exponents, $a^0= 1[\because$ a is non-zero integer$]$
$\therefore (-2)^0= 1$
$LHS \neq RHS$
View full question & answer→Question 931 Mark
If $36 = 6 × 6 = 6^2$, then $\frac{1}{36}$ expressed as a power with the base $6$ is ________.
AnswerGiven,
$36 = 6 \times 6 = 6^2$
So,
$\frac{1}{36}=\frac{1}{6\times6}=\frac{1}{(6)^2}=(6)^{-2}$
$\Big[\because\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}\Big]$
View full question & answer→Question 941 Mark
$\text{a}^\text{m}=\frac{1}{\text{a}^{-\text{m}}}$
AnswerUsing law of exponents, $\text{a}^\text{m}=\frac{1}{\text{a}^{-\text{m}}}$
$LHS = RHS$
View full question & answer→Question 951 Mark
The multiplicative inverse of $10^{10}$ is ___________.
AnswerFor multiplicative inverse, a is called the multiplicative inverse of $b,$ if $a \times b = 1$
Put $b = 10^{10}$
Then,
$a \times 10^{10}= 1$
$\Rightarrow\text{a}=\frac{1}{10}^{10}\ \Big[\because\frac{1}{\text{a}^\text{m}}=\text{a}^{-\text{m}}\Big]$
$\therefore\text{a}=10^{-10}$
Hence,
The multiplicative inverse of $10^{10}$ is $10^{-10}$.
View full question & answer→