MCQ 511 Mark
The value of $(7^{-1}- 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1}$ is:
AnswerA. $44$
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^{\text{m}}}$ [$\therefore$ a is non-zero integer]
$\therefore$ $(7^{-1}- 8^{-1})^{-1} - (3^{-1} - 4^{-1})^{-1}$
$=\Big(\frac{1}{7}-\frac{1}{8}\Big)^{-1}-\Big(\frac{1}{3}-\frac{1}{4}\Big)^{-1}$
$=\Big(\frac{1}{56}\Big)^{-1}-\Big(\frac{1}{12}\Big)^{-1}$
$=56-12=44$
View full question & answer→MCQ 521 Mark
The multiplicative inverse of $7^{-2}$is:
- A
$7$
- ✓
$7^2$
- C
$\frac{1}{7^2}$
- D
$\frac{1}{7}$
AnswerB. $7^2$
Solution:
The multiplicative inverse of any value is the one which when multiplied by the original value gives a value equal to $1.$
$7^{-2} = \frac{1}{7^2}$
Hence, $7^{-2}\times\frac{1}{7^2} = 1$
View full question & answer→MCQ 531 Mark
Tick $(\checkmark)$ the correct answer the following:
The value of $(-2)^{-5}$ is-
- A
$-32$
- ✓
$\frac{-1}{32}$
- C
$32$
- D
$\frac{1}{32}$
AnswerCorrect option: B. $\frac{-1}{32}$
B. $\frac{-1}{32}$
Solution:
$(-2)^{-5}$
$=\frac{1}{(-2)^5}$
$=\frac{1}{-32}$
$=\frac{-1}{32}\Big[\because(\text{x})^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big]$
View full question & answer→MCQ 541 Mark
Tick $(\checkmark)$ the correct answer the following:
$\bigg\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}\div\Big(\frac{1}{4}\Big)^{-3}=\ ?$
- ✓
$\frac{19}{64}$
- B
$\frac{27}{16}$
- C
$\frac{64}{19}$
- D
$\frac{16}{25}$
AnswerCorrect option: A. $\frac{19}{64}$
$\bigg\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\bigg\}\div\Big(\frac{1}{4}\Big)^{-3}$
$=\big(3^3-2^3\big)\div(4)^3$
$=(27-8)\div64$
$=\frac{19}{64}$
View full question & answer→MCQ 551 Mark
The value of $2^{-2}$ is:
- A
$4$
- ✓
$\frac{1}{4}$
- C
$2$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\frac{1}{4}$
B. $\frac{1}{4}$
Solution:
$2^2 = \frac{1}{2^2} = \frac{1}{4}$
View full question & answer→MCQ 561 Mark
$(a^m)^n$ is equal to:
- A
$a^{m+n}$
- B
$a^{m-n}$
- ✓
$a^{mn}$
- D
$a^{n-m}$
AnswerCorrect option: C. $a^{mn}$
C. $a^{mn}$
View full question & answer→MCQ 571 Mark
For any two non$-$zero rational nmbers a, $(\text{a}^3)^{-2}$ is equal to:
- A
$\text{a}^9$
- ✓
$\text{a}^{-6}$
- C
$\text{a}^{-9}$
- D
$\text{a}^1$
AnswerCorrect option: B. $\text{a}^{-6}$
$(\text{a})^{-2}=\text{a}^3\times(-2)$
$=\text{a}^{-6}$
View full question & answer→MCQ 581 Mark
For a non-zero rational number $z, (z^{-2})^3$ is equal to:
- A
$z^6$
- ✓
$z^{-6}$
- C
$z^1$
- D
$z^4$
AnswerCorrect option: B. $z^{-6}$
B. $z^{-6}$
Solution:
Using law of exponents, $(a^m)^n = (a)^{mn}$ [$\because$ a is non-zero integer]
Similarly,
$(z^{-2})^3 = (z)^{(-2) \times 3}$
$= (z)^{-6}$
View full question & answer→MCQ 591 Mark
$2^{\text{log}^3_2} + 3^{\text{log}^2_3}$ is equal to:
View full question & answer→MCQ 601 Mark
Which of the following is the multiplicative inverse of $(3 \times 4)^{-2}?$
- A
$\frac{1}{144}$
- B
$144$
- C
$\frac{1}{12}$
- ✓
$12$
View full question & answer→MCQ 611 Mark
The value of $\Big(-\frac{2}{3}\Big)^4$ is equal to:
- ✓
$\frac{16}{81}$
- B
$\frac{81}{16}$
- C
$\frac{-16}{81}$
- D
$\frac{81}{-16}$
AnswerCorrect option: A. $\frac{16}{81}$
A. $\frac{16}{81}$
Solution:
Given,
$\Big(\frac{-2}{3}\Big)^4$
$=\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big)$
$=\frac{16}{81}$
[for $(-a)^m,$ if m is even, then $(-a)^m$ is positive]
View full question & answer→MCQ 621 Mark
Find the value of x, if $32 = 2^x.$
View full question & answer→MCQ 631 Mark
Tick $(\checkmark)$ the correct answer the following:
The value of x for which $\Big(\frac{7}{12}\Big)^{-4}\times\Big(\frac{7}{12}\Big)^{3\text{x}}=\Big(\frac{7}{12}\Big)^5$, is:
Answer$\Big(\frac{7}{12}\Big)^{-4}\times\Big(\frac{7}{12}\Big)^{3\text{x}}=\Big(\frac{7}{12}\Big)^5$
$\Rightarrow\Big(\frac{7}{12}\Big)^{3\text{x}-4}$
$=\Big(\frac{7}{12}\Big)^5$
$=3\text{x}-4=5$
$=3\text{x}=5+4=9$
$\Rightarrow\text{x}=\frac{9}{3}$
$=3$
View full question & answer→MCQ 641 Mark
If $1$ nanometer is equal to $\frac{1}{1000000000}\text{m}$ Write $23$ nanometer in meter and in standard form:
AnswerCorrect option: A. $2.3 \times 10^{-8}m$
A. $2.3 \times 10^{-8}m$
Solution:
$1\text{nm}=\frac{1}{1000000000}\text{m}=1\times10^{-9}$
Multiplying $23$ to both side,
$23nm = 23 \times 1 \times 10^{-9}$
$= 23 \times 10^{-9}m$
$= 23 \times 10 \times 10^{-9}$
$=2.3 \times 10^{-8}m$
View full question & answer→MCQ 651 Mark
The usual form for $2.03 × 10^{-5}$
- A
$0.203$
- B
$0.00203$
- C
$203000$
- ✓
$0.0000203$
AnswerCorrect option: D. $0.0000203$
D. $0.0000203$
Solution:
Given,
$2.03 × 10^{-5} = 0.0000203$
[$\therefore$ placing decimal five digit towards left of original position]
View full question & answer→MCQ 661 Mark
$\Big(\frac{2}{3}\Big)^{-5}\times\Big(\frac{5}{7}\Big)^{-5}$ is equal to:
- A
$\Big(\frac{2}{3}\times\frac{5}{7}\Big)^{-10}$
- ✓
$\Big(\frac{2}{3}\times\frac{5}{7}\Big )^{-5}$
- C
$\Big(\frac{2}{3}\times\frac{5}{7}\Big)^{25}$
- D
$\Big(\frac{2}{3}\times\frac{5}{7}\Big)^{-25}$
AnswerCorrect option: B. $\Big(\frac{2}{3}\times\frac{5}{7}\Big )^{-5}$
We have:
$\Big(\frac{2}{3}\Big)^{-5}\times\Big(\frac{5}{7}\Big)^{-5}=\Big(\frac{2}{3}\times\frac{5}{7}\Big)^{-5}$
View full question & answer→MCQ 671 Mark
Write the expression using exponents: $61 \times 61 \times 61 \times 61 \times 61.$
- A
$6^{12}$
- B
$6^{13}$
- C
$6^{14}$
- ✓
$6^{15}$
AnswerCorrect option: D. $6^{15}$
D. $6^{15}$
View full question & answer→MCQ 681 Mark
Which of the following number is not equal to $\frac{-8}{27}?$
- ✓
$\Big(\frac{2}{3}\Big)^{-3}$
- B
$-\Big(\frac{2}{3}\Big)^3$
- C
$\Big(-\frac{2}{3}\Big)^3$
- D
$\Big(\frac{-2}{3}\Big)\times\Big(\frac{-2}{3}\Big )\times\Big(\frac{-2}{3}\Big)$
AnswerCorrect option: A. $\Big(\frac{2}{3}\Big)^{-3}$
We can write $\frac{-8}{27}$ as $\frac{-2\times(-2)\times(-2)}{3\times3\times3}$ it can be written in the foms given below.
$\frac{-2\times(-2)\times(-2)}{3\times3\times3}=\frac{2\times2\times2}{3\times3\times3 }$
$=-\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$
$=-\Big(\frac{2}{3}\Big)^3$
View full question & answer→MCQ 691 Mark
The multiplicative inverse of $7^{-2}$is:
- ✓
$7^2$
- B
$7$
- C
$\frac{1}{7^2}$
- D
$\frac{1}{7}$
AnswerA. $7^2$
Solution:
The multiplicative inverse of any value is the one which when multiplied by the original value gives a value equal to $1.$
$7^{-2} = \frac{1}{7^2}$
Hence, $7^{2}\times\frac{1}{7^2}=1$
View full question & answer→MCQ 701 Mark
$0.00001275$ is equal to:
- ✓
$1.275 \times 10^{-5}$
- B
$1.275 \times 10^{-3}$
- C
$1.275 \times 10^{-4}$
- D
$1.275 \times 10^{3}$
AnswerCorrect option: A. $1.275 \times 10^{-5}$
A. $1.275 \times 10^{-5}$
Solution:
$0.00001275 = 1.275 \times 10^{-5}$
View full question & answer→MCQ 711 Mark
Write $0.00000000256$ in standard form:
- A
$2.56 \times 10^{-11}$
- B
$2.56 \times 10^{-10}$
- C
$2.56 \times 10^{-8}$
- ✓
$2.56 \times 10^{-9}$
AnswerCorrect option: D. $2.56 \times 10^{-9}$
D. $2.56 \times 10^{-9}$
Solution:
$0.00000000256 =\frac{256}{100000000000}$
$=\frac{2.56\times100}{100000000000}$
$=\frac{2.56}{100000000000}$
$=2.56\times10^{-9}$
View full question & answer→MCQ 721 Mark
Tick $(\checkmark)$ the correct answer the following:
$3670000$ in standard form is:
- A
$367 \times 10^4$
- B
$36.7 \times 10^5$
- ✓
$3.67 \times 10^6$
- D
AnswerCorrect option: C. $3.67 \times 10^6$
C. $3.67 \times 10^6$
Solution:
$3670000 = 3.670000 \times 1000000$
$=3.67 \times 10^6$
View full question & answer→MCQ 731 Mark
If x be any integer different from zero and m, n be any integers, then $(x^m)^n$ is equal to:
- A
$\text{x}^{\text{m}+\text{n}}$
- ✓
$\text{x}^{\text{mn}}$
- C
$\text{x}^{\frac{\text{m}}{\text{n}}}$
- D
$\text{x}^{\text{m}-\text{n}}$
AnswerCorrect option: B. $\text{x}^{\text{mn}}$
B. $\text{x}^{\text{mn}}$
Solution:
Using law of exponents, $(\text{a}^{\text{m}})^{\text{n}}=(\text{a})^{\text{m}\times\text{n}}$ [$\because$ a is non-zero integer]
Similaly,
$(\text{x}^{\text{m}})^{\text{n}}=(\text{x})^{\text{m}\times\text{n}}$
$=(\text{x})^\text{mn}$
View full question & answer→MCQ 741 Mark
Simplify: $4^3 \div 4^{-7}$
- A
$4^8$
- ✓
$4^{10}$
- C
$4^4$
- D
$4^{21}$
AnswerCorrect option: B. $4^{10}$
B. $4^{10}$
Solution:
$4^3 \div 4^{-7} = 4^{(3-(-7))} (a^m \div a^n = a^{m-n})$
$= (4)^{(3+7)} = 4^{10}$
View full question & answer→MCQ 751 Mark
$865000$ is equal to:
- ✓
$8.65 \times 10^5$
- B
$8.65 \times 10^3$
- C
$8.65 \times 10^6$
- D
$8.65 \times 10^4$
AnswerCorrect option: A. $8.65 \times 10^5$
A. $8.65 \times 10^5$
View full question & answer→MCQ 761 Mark
The multiplicative inverse of $10^5$ is:
- A
$5$
- B
$10$
- ✓
$10^{-5}$
- D
$10^5$
AnswerCorrect option: C. $10^{-5}$
C. $10^{-5}$
Solution:
$10^5× 10^{-5} = 2^{5-5} = 10^\circ = 1$
View full question & answer→MCQ 771 Mark
The value of $3^\circ$is ________.
View full question & answer→MCQ 781 Mark
Tick $(\checkmark)$ the correct answer the following : $\Big(\frac{-3}{4}\Big)^{2}=\ ?$
- A
$\frac{-9}{16}$
- ✓
$\frac{9}{16}$
- C
$\frac{16}{9}$
- D
$\frac{-16}{9}$
AnswerCorrect option: B. $\frac{9}{16}$
$\Big(\frac{-3}{4}\Big)^{2}$
$=\Big(\frac{-3}{4}\Big)\times\Big(\frac{-3}{4}\Big)$
$=\frac{9}{16}$
View full question & answer→MCQ 791 Mark
Mark $(\checkmark)$ against the correct answer of the following : The value of $\Big(\frac{3}{4}\Big)^{-3}$ is:
- A
$\frac{-27}{64}$
- ✓
$\frac{64}{27}$
- C
$\frac{-9}{4}$
- D
$\frac{27}{64}$
AnswerCorrect option: B. $\frac{64}{27}$
$=\Big(\frac{3}{4}\Big)^{-3}$
$=\Big(\frac{4}{3}\Big)^{3}$
$=\frac{4^3}{3^3}$
$=\frac{64}{27}$
View full question & answer→MCQ 801 Mark
The expression, $(5^{-1} + 7^{-1} + 3^{-1})^0$ is equals to.
- A
$15^{-3}$
- B
$-3$
- C
$15^{-1}$
- ✓
$1$
View full question & answer→MCQ 811 Mark
Which of the following is used as a form of $5.05\times10^6$?
- A
$505000$
- B
$505000000$
- ✓
$5050000$
- D
$50500000$
AnswerCorrect option: C. $5050000$
$5.05\times10^6$
$=5.05\times1000000=5050000$
View full question & answer→MCQ 821 Mark
Write $23569874500$ in standard form:
- A
$2.35698745 \times 10^9$
- ✓
$2.35698745 \times 10^{10}$
- C
$2.35698745 \times 10^{8}$
- D
$2.35698745 \times 10^{11}$
AnswerCorrect option: B. $2.35698745 \times 10^{10}$
B. $2.35698745 \times 10^{10}$
Solution:
$23569874500 = 235698745 \times 100$
$= 2.35698745 \times 10^8 \times 10^2$
$= 2.35698745 \times 10^{10}$
View full question & answer→MCQ 831 Mark
Find the multiplicative inverse of $7^{-2}.$
View full question & answer→MCQ 841 Mark
The value of $(3^4)^3$ is:
AnswerCorrect option: B. $3^{12}$
B. $3^{12}$
Solution:
By law of exponent:
$(a^m)^n = a^{mn}$
$(3^4)^3 = 3^{4 \times 3}$
$= 3^{12}$
View full question & answer→MCQ 851 Mark
Find the value of $\Big(\frac{3}{2}\Big)^{-3}\times2^{-5}$
- ✓
$\frac{1}{108}$
- B
$\frac{1}{18}$
- C
$\frac{1}{31}$
- D
$108$
AnswerCorrect option: A. $\frac{1}{108}$
$\Big(\frac{3}{2}\Big)^{-3}\times2^{-5}=\frac{3^{-3}}{2^{-5}}\times\frac{1}{2^5}$
$=\frac{2^{3}}{3^{3}}\times\frac{1}{2^5}$
$=\frac{1}{3^{3}}\times\frac{1}{2^2}$
$=\frac{1}{27}\times\frac{1}{4}$
$=\frac{1}{108}$
View full question & answer→MCQ 861 Mark
$\Big(-\frac{5}{7}\Big)^{-5}$ is equal to:
- A
$\Big(-\frac{5}{7}\Big)^{-5}$
- B
$\Big(\frac{5}{7}\Big)^5$
- C
$\Big(\frac{7}{5}\Big)^5$
- ✓
$\Big(\frac{-7}{5}\Big)^5$
AnswerCorrect option: D. $\Big(\frac{-7}{5}\Big)^5$
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}} [\because$ a is non$-$zero integer$]$
$\therefore$ $\Big(-\frac{5}{7}\Big)^5=\frac{1}{\Big(\frac{-5}{7}\Big)^5}$
$=\Big(-\frac{7}{5}\Big)^5$
View full question & answer→MCQ 871 Mark
$(-2)^{-2}$ is equal to:
- ✓
$\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$-\frac{1}{2}$
- D
$-\frac{1}{4}$
AnswerCorrect option: A. $\frac{1}{4}$
A. $\frac{1}{4}$
Solution:
$(-2)^{-2}=\frac{1}{(-2)^2} = \frac{1}{4}$
View full question & answer→MCQ 881 Mark
$695000$ is equal to:
- ✓
$6.95 \times 10^5$
- B
$6.95 \times 10^3$
- C
$6.95 \times 10^6$
- D
$6.95 \times 10^4$
AnswerCorrect option: A. $6.95 \times 10^5$
A. $6.95 \times 10^5$
Solution:
$695000 = 6.95 \times 10^5$
View full question & answer→MCQ 891 Mark
The value of $\log^{\text{a}^{2}}_\text{abc}+\log^{\text{b}^{2}}_\text{abc}+\log^{\text{c}^{2}}_\text{abc}$ is equal to:
View full question & answer→MCQ 901 Mark
The standard form of $4050000$ is given by.
- ✓
$4.05 \times 10^6$
- B
$4.05 \times 10^9$
- C
$405 \times 10^6$
- D
$4.05 \times 10^{-6}$
AnswerCorrect option: A. $4.05 \times 10^6$
A. $4.05 \times 10^6$
View full question & answer→MCQ 911 Mark
For a non-zero integer x, $(x^4)^{–3}$ is equal to:
- A
$x^{12}$
- ✓
$x^{-12}$
- C
$x^{64}$
- D
$x^{-64}$
AnswerCorrect option: B. $x^{-12}$
B. $x^{-12}$
Solution:
Using law of exponents, $(a^m)^n = (a)^{m \times n} = (a^m)^n$ [$\because$ a is non-zero integer]
similarly,
$(x^4)^{-3} = (x)^{4 \times (-3)}$
$= x^{-12}$
View full question & answer→MCQ 921 Mark
$\text{log}^\text{yz}_\text{xy}\times\text{log}^\text{zx}_\text{yz}\times\text{log}^\text{xy}_\text{zx}$ is equal to:
View full question & answer→MCQ 931 Mark
For a non-zero integer $x, x^7 \div x^{12}$ is equal to:
- A
$x^{5}$
- B
$x^{19}$
- ✓
$x^{-5}$
- D
$x^{-19}$
AnswerCorrect option: C. $x^{-5}$
C. $x^{-5}$
View full question & answer→MCQ 941 Mark
If x be any non-zero integer, then $x{-1}$ is equal to:
- A
$\text{x}$
- ✓
$\frac{1}{\text{x}}$
- C
$-\text{x}$
- D
$\frac{-1}{\text{x}}$
AnswerCorrect option: B. $\frac{1}{\text{x}}$
B. $\frac{1}{\text{x}}$
Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{a^{\text{m}}}$ [$\because$ a is non-zero integer]
Similarly,
$\text{x}^{-1}=\frac{1}{\text{x}}$
View full question & answer→MCQ 951 Mark
$\Big(\frac{-1}{2}\Big)^5\times\Big(\frac{-1}{2}\Big)^3$ is equal to:
- ✓
$\Big(\frac{-1}{2}\Big)^{8}$
- B
$-\Big(\frac{1}{2}\Big)^8$
- C
$\Big(\frac{1}{4}\Big)^8$
- D
$\Big(-\frac{1}{2}\Big)^{15}$
AnswerCorrect option: A. $\Big(\frac{-1}{2}\Big)^{8}$
We have:
$\Big(\frac{-1}{2}\Big)^5\times\Big(\frac{-1}{2}\Big)^3$
$=\Big(\frac{-1}{2}\Big)^{5+3}$
$=\Big(\frac{-1}{2}\Big)^8$
View full question & answer→MCQ 961 Mark
The value of $10000$ is.
View full question & answer→MCQ 971 Mark
For any two non$-$zero rational nmbers $a$ and $b, \text{a}^{4}\div\text{b}^4$ is equal to:
- A
$(\text{a}\div\text{b})^1$
- B
$(\text{a}\div \text{b})^0$
- ✓
$(\text{a}\div \text{b})^4$
- D
$(\text{a}\div \text{b})^8$
AnswerCorrect option: C. $(\text{a}\div \text{b})^4$
This is one of the basic exponential formulae,
i.e. $(\text{a}\div\text{b})^\text{n}=\text{a}^\text{n}\div\text{b}^\text{n}$
View full question & answer→MCQ 981 Mark
Which of the following is the value of 'm' in $\frac{6^{\text{m}}}{6^{-3}}=6^{5}$?
AnswerThe question given to us is:
$\frac{6^{\text{m}}}{6^{-3}}=6^{5}$
This can be re-written as:
$6^{\text{m}-(-3)}=6^{5}\big(\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}\big)$
Thus, we get:
$6^{\text{m+3}}=6^{5}$
Now, comparing the exponents in the above equation, we will get:
$\text{m}+3=5$
$\therefore\text{m}=5-3=2$
Thus, the last option, option IV, $m = 2$, is the correct option.
View full question & answer→MCQ 991 Mark
If $\log^7_{10}=0.81$ and $\log^2_{10}=0.30$ then $\log^{49}_4$ is equal to:
View full question & answer→MCQ 1001 Mark
$\Big(\frac{1}{5}\Big)^0$ is equal to:
- A
$0$
- B
$\frac{1}{5}$
- ✓
$1$
- D
$5$
AnswerWe have:
$\Big(\frac{1}{5}\Big)^0=1$
View full question & answer→