Question 14 Marks
If $3^{4 x}=(81)^{-1}$ and $10^{\frac{1}{y}}=0.0001$, Find the value of $2^{-x} \times 16^y$
Answer$ 3^{4 x}=(81)^{-1}$ and $10^{\frac{1}{y}}=0.0001$
$ \Rightarrow 3^{4 x}=\left(3^4\right)^{-1}$ and $10^{\frac{1}{y}}=\frac{1}{10000}$
$ \Rightarrow 3^{4 x}=3^{-4}$ and $10^{\frac{1}{y}}=\frac{1}{10^4}$
$ \Rightarrow 4 x=-4$ and $10^{\frac{1}{y}}=10^{-4}$
$ \Rightarrow x=-1$ and $\frac{1}{y}=-4$
$ \Rightarrow x=-1$ and $y=-\frac{1}{4}$
$ \therefore 2^{-x} \times 16^y$
$=2^{-(-1)} \times 16^{-\frac{1}{4}}$
$ =2 \times 2^{4 \times-\frac{1}{4}}$
$ =2 \times 2^{-1}$
$ =2^{1-1}$
$ =2^0$
$ =1$
View full question & answer→Question 24 Marks
If $2^{\mathrm{x}}=4^{\mathrm{y}}=8^{\mathrm{z}}$ and $\frac{1}{2 x}+\frac{1}{4 y}+\frac{1}{8 z}=4$, find the value of $\mathrm{x}$.
Answer$2^{\mathrm{x}}=4^{\mathrm{y}}=8^{\mathrm{z}}$ and $\frac{1}{2 x}+\frac{1}{4 y}+\frac{1}{8 z}=4$
$2^x=4^y=8^z$
$ \Rightarrow 2^x=2^{2 y}=2^{3 z}$
$ \Rightarrow x=2 y=3 z$
$\Rightarrow \mathrm{y}=\frac{x}{2}$ and $z=\frac{x}{3}$
Now, $\frac{1}{2 x}+\frac{1}{4 y}+\frac{1}{8 z}=4$
$\Rightarrow \frac{1}{2 x}+\frac{1}{\frac{4 x}{2}}+\frac{1}{\frac{8 x}{3}}=4$
$ \Rightarrow \frac{1}{2 x}+\frac{2}{4 x}+\frac{3}{8 x}=4$
$\Rightarrow \frac{1}{2 x}+\frac{1}{2 x}+\frac{3}{8 x}=4$
$\Rightarrow \frac{4+4+3}{8 x}=4$
$\Rightarrow \frac{11}{8 x}=4$
$\Rightarrow x=\frac{11}{32}$
View full question & answer→Question 34 Marks
Prove that : $\frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}=\frac{2}{b^2-a^2}$
Answer$ \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}=\frac{2}{b^2-a^2}$
$ \text { L.H.S. }$
$=\frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}$
$ =\frac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}}+\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}$
$ =\frac{\frac{1}{a}}{\frac{b+a}{a b}}+\frac{\frac{1}{a}}{\frac{b-a}{a b}}$
$ =\frac{1}{a} \times \frac{a b}{b+a}+\frac{1}{a} \times \frac{a b}{b-a}$
$ =\frac{b}{b+a}+\frac{b}{b-a}$
$ =\frac{b^2-a b+b^2+a b}{b^2-a^2}$
$ =\frac{2 b^2}{b^2-a^2}$
$ =\text { R.H.S. }$
View full question & answer→Question 44 Marks
If $m=\sqrt[3]{15}$ and $n=\sqrt[3]{14}$, find the value of $m-n-\frac{1}{m^2+m n+n^2}$
Answer$ \sqrt[3]{15}$ and $n=\sqrt[3]{14}$
$ \Rightarrow \mathrm{m}^3=15$ and $\mathrm{n}^3=14$
$ \therefore \mathrm{m}-\mathrm{n}-\frac{1}{m^2+m n+n^2}$
$ =\frac{\left(m^3+m^2 n+m n^2\right)-\left(m^2 n+m n^2+n^3\right)-1}{m^2+m n+n^2}$
$ =\frac{m^3+m^2 n+m n^2-m^2 n-m n^2-n^3-1}{m^2+m n+n^2}$
$ =\frac{m^3-n^3-1}{m^2+m n+n^2}$
$ =\frac{15-14-1}{m^2+m n+n^2}$
$ =\frac{1-1}{m^2+m n+n^2}$
$ =0$
View full question & answer→Question 54 Marks
Evaluate : $(64)^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+(27)^{-\frac{2}{3}} \times\left(\frac{25}{9}\right)^{-\frac{1}{2}} \mathrm{}$
Answer$ (64)^{\frac{2}{3}}-\sqrt[3]{125}-\frac{1}{2^{-5}}+(27)^{-\frac{2}{3}} \times\left(\frac{25}{9}\right)^{-\frac{1}{2}}$
$ =\left(4^3\right)^{\frac{2}{3}}-\sqrt[3]{5^3}-2^5+\left(3^3\right)^{-\frac{2}{3}} \times\left(\frac{5^2}{3^2}\right)^{-\frac{1}{2}}$
$ =4^2-5-2^5+3^{-2} \times\left(\frac{5}{3}\right)^{2 \times\left(-\frac{1}{2}\right)}$
$ =16-5-32+\frac{1}{3^2} \times\left(\frac{5}{3}\right)^{-1}$
$ =-21+\frac{1}{9} \times \frac{3}{5}$
$ =-21+\frac{1}{15}$
$ =\frac{-315+1}{15}$
$ =-\frac{314}{15}$
$ =-20 \frac{14}{15}$
View full question & answer→Question 64 Marks
Solve :$2^{2x}+ 2^{x+2} - 4 \times 2^3= 0$
Answer$2^{2x}+ 2^{x+2} - 4 \times 2^3= 0$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 4 \times 2 \times 2 \times 2 = 0$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 32 = 0$
Putting $y = 2^x$
$\Rightarrow y^2 + 4y - 32 = 0$
$\Rightarrow y2 + 8y - 4y - 32 = 0$
$\Rightarrow y( y + 8 ) - 4( y + 8 ) = 0$
$\Rightarrow ( y + 8 )( y - 4 ) = 0$
$\Rightarrow y + 8 = 0$ or $y - 4 = 0$
$\Rightarrow y = - 8$ or $y = 4$
$\Rightarrow 2^x = - 8$ or $2^x = 4$
$\Rightarrow 2^x = 2^2$
$[ \because 2^x = - 8$ is not possible. $]$
$\Rightarrow x = 2.$
View full question & answer→Question 74 Marks
Solve : $\left[3^x\right]^2: 3 \mathrm{x}=9: 1$
Answer
$ {\left[3^x\right]^2: 3 x=9: 1}$
$ \Rightarrow \frac{\left[3^x\right]^2}{3^x}=\frac{9}{1}$
$ \Rightarrow\left[3^x\right]^2=9 \times 3^x$
$ \Rightarrow\left[3^x\right]^2=3^2 \times 3^x$
$ \Rightarrow\left[3^x\right]^2=3^{x+2}$
We know that if bases are equal, the powers are equal.
$\Rightarrow x^2=x+2$
$ \Rightarrow x^2-x-2=0$
$ \Rightarrow x^2-2 x+x-2=0$
$ \Rightarrow x(x-2)+1(x-2)=0$
$ \Rightarrow(x+1)(x-2)=0$
$ \Rightarrow x+1=0$ or $x-2=0$
$ \Rightarrow x=-1$ or $x=2$
View full question & answer→Question 84 Marks
Find $x$, if : $\left(\sqrt[3]{\frac{2}{3}}\right)^{x-1}=\frac{27}{8}$
Answer$\left(\sqrt[3]{\frac{2}{3}}\right)^{x-1}=\frac{27}{8}$
$ {\left[\left(\frac{2}{3}\right)^{\frac{1}{3}}\right]^{x-1}=\frac{3^3}{2^3}}$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{3}{2}\right)^3$
$ \Rightarrow\left(\frac{2}{3}\right)^{\frac{x-1}{3}}=\left(\frac{2}{3}\right)^{-3}$
We know that if bases are equal, the powers are equal
$\Rightarrow \frac{x-1}{3}=-3$
$ \Rightarrow x-1=-9$
$ \Rightarrow x=-9+1$
$ \Rightarrow x=-8$
View full question & answer→Question 94 Marks
Find $x$, if : $\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$
Answer$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$
$ \Rightarrow\left[\left(\frac{3}{5}\right)^{\frac{1}{2}}\right]^{x+1}=\frac{5 \times 5 \times 5}{3 \times 3 \times 3}$
$\Rightarrow\left(\frac{3}{5}\right)^{\frac{x+1}{2}}=\left(\frac{5}{3}\right)^3$
$\Rightarrow\left(\frac{3}{5}\right)^{\frac{x+1}{2}}=\left(\frac{3}{5}\right)^{-} 3$
We know that if bases are equal, the powers are equal
$\Rightarrow \frac{x+1}{2}=-3$
$\Rightarrow x+1=-6$
$ \Rightarrow x=-6-1$
$ \Rightarrow x=-7$
View full question & answer→Question 104 Marks
If $4^{x + 3}= 112 + 8 \times 4^x,$ find the value of $(18x)^{3x}.$
Answer$4^{x+3}=112+8 \times 4^x$
$\Rightarrow 4^x \times 4^3=112+8 \times 4^x$
$\Rightarrow 64 \times 4^x=112+8 \times 4^x$
Let $4^x=y$
$64 y=112+8 y$
$\Rightarrow 56 y=112$
$\Rightarrow y=2$
Substituting We get,
$4^x=2$
$\Rightarrow 2^{2 x}=2$
$\Rightarrow 2 x=1$
$\Rightarrow x=\frac{1}{2}$
$(18 x)^{3 x}=\left(\frac{18}{2}\right)^{3 \times \frac{1}{2}}$
$=9^{3 \times \frac{1}{2}}=\left(9^{\frac{1}{2}}\right)^3$
$=3^3=27$
View full question & answer→Question 114 Marks
If $m \neq n$ and $(m + n)^{-1}(m^{-1}+ n^{-1}) =m^xn^y$, show that : $x + y + 2 = 0$
Answer$ (\mathrm{m}+\mathrm{n})^{-1}\left(\mathrm{~m}^{-1}+\mathrm{n}^{-1}\right)=\mathrm{m}^{\mathrm{x}} \mathrm{n}^{\mathrm{y}}$
$ \Rightarrow \frac{1}{m+n} \times\left(\frac{1}{m}+\frac{1}{n}\right)=m^x \cdot n^y$
$ \Rightarrow \frac{1}{\not m+ \not n} \times\left(\frac{\not m+\not n}{m n}\right)=m^x \cdot n^y$
$ \Rightarrow \frac{1}{m n}=m^x \cdot n^y$
$ \Rightarrow m^{-1} n^{-1}=m^x \cdot n^y$
Comparing the coefficient of $x$ and $y$, we get
$x=-1$ and $y=-1$
Putting $x=-1$ and $y=-1$
$x+y+2$
$-1+(-1)+2$
$-1-1+2$
$=0 \text { L.H.S }$
View full question & answer→Question 124 Marks
If $a =x^{m+ n}.y^l; b =x^{n+l}.y^m$ and $c = x^{l+ m}.y^n,$Prove that : $a^{m - n}.b^{n-l}.c^{l- m}= 1$
Answer$ a=x^{m+n} \cdot y^{\prime}$
$b=x^{n+1} \cdot y^m$
$c=x^{l+m} \cdot y^n$
$a^{m-n} \cdot b^{n-I} \cdot c^{1-m}$
$=1\text { LHS }$
$=a^{m-n} \cdot b^{n-1} \cdot c^{1-m}$
$=\left[x^{(m+n) \cdot} \cdot y^1\right]^{(m-n)} \cdot\left[x^{(n+1)} \cdot y^m\right]^{(n-1)} \cdot\left[x^{(1+m)} \cdot y^n\right]^{(1-m)}$
$=x^{(m+n)(m-n)} \cdot y^{(m-n)} \cdot x^{(n+1)(n-1)} \cdot y^{m(n-1)} \cdot x^{(1+m)(1-m)} .$
$y^{n(I-m)}$
$=x^{m^2-n^2+n^2-l^2+l^2-m^2} \cdot y^{l m-\ln +m n-m l+n l-n m}$
$=x^0 \cdot y^0$
$=1 \text{RHS}$
View full question & answer→Question 134 Marks
If $1960 = 2^a. 5^b. 7^c$, calculate the value of $2^{-a}. 7^b. 5^{-c}.$
Answer$1960=2^a \times 5^b \times 7^c$
$ \Rightarrow 2 \times 2 \times 2 \times 5 \times 7 \times 7=2^a \times 5^b \times 7^c$
$ \Rightarrow 2^3 \times 5^1 \times 7^2=2^a \times 5^b \times 7^c$
$\Rightarrow 2^a \times 5^b \times 7^c=2^3 \times 5^1 \times 7^2$
Comparing powers of 2,5 and 7 on the both sides of equation,
We have
$a=3 ; b=1$ and $c=2$
Hence,
Value of $2^{-a} \times 7^b \times 5^{-c}$
$=2^{-3} \times 7^1 \times 5^{-2}$
$=\frac{1}{2^3} \times 7 \times \frac{1}{5^2}$
$=\frac{1}{8} \times 7 \times \frac{1}{5 \times 5}$
$=\frac{7}{200}$
View full question & answer→Question 144 Marks
Evaluate:$\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}$
Answer$ \sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}$
$ =\sqrt{\frac{1}{2} \times \frac{1}{2}}+(0.1 \times 0.1)^{-\frac{1}{2}}-(3 \times 3 \times 3)^{\frac{2}{3}}$
$ =\frac{1}{2}+\left[(0.1)^2\right]^{-\frac{1}{2}}-\left(3^2\right)^{\frac{2}{3}}$
$ =\frac{1}{2}+(0.1)^{2 \times\left(-\frac{1}{2}\right)}-3 \times\left(3 \times \frac{2}{3}\right)$
$ =\frac{1}{2}+(0.1)^{-1}-3^2$
$ =\frac{1}{2}+\frac{1}{0.1}-9$
$ =\frac{1}{2}+\frac{10}{1}-9$
$ =\frac{1+20-18}{2}$
$ =\frac{3}{2}$
$ =1 \frac{1}{2}$
View full question & answer→Question 154 Marks
Solve $x$ and $y$ if : $( \sqrt{32} )^x \div 2^{y + 1} = 1$ and $8^y - 16^{4 - x/2} = 0$
AnswerConsider the quation
$(\sqrt{32})^x \div 2^{y+1}=1$
$ \Rightarrow(\sqrt{2 \times 2 \times 2 \times 2 \times 2})^x \div 2^{y+1}=1$
$ \Rightarrow\left(\sqrt{2^5}\right)^x \div 2^{y+1}=1$
$ \Rightarrow\left[\left(2^5\right)^{\frac{1}{2}}\right]^x \div 2^{y+1}=x^0$
$ \Rightarrow 2^{5 \frac{x}{2}} \div 2^{y+1}=x^0$
$ \Rightarrow \frac{5 x}{2}-(y+1)=0$
$ \Rightarrow 5 \mathrm{x}-2(y+1)=0$
$\Rightarrow 5 x-2 y-2=0\ldots(1)$
Now consider the other equation
$8^y-16^{4-\frac{x}{2}}=0$
$ \Rightarrow\left(2^3\right)^y-\left(2^4\right)^{4-\frac{x}{2}}=0$
$ \Rightarrow 2^{3 y}-2^{4\left(4-\frac{x}{2}\right)}=0$
$ \Rightarrow 2^{3 y}=2^{4\left(4-\frac{x}{2}\right)}$
$ \Rightarrow 3 y=4\left(4-\frac{x}{2}\right)$
$ \Rightarrow 3 y=16-2 x$
$\Rightarrow 2 x+3 y=16 \ldots(2)$
Thus, We have two equations,
$5 x-2 y=2\ldots(1)$
$2 x+3 y=16\ldots(2)$
Multiplying equation $(1)$ by $3$ and $(2)$ by $2 ,$ We have
$15 x-6 y=6\ldots .(3)$
$4 x+6 y=32\ldots .(4)$
Adding equation $(3)$ and $(4),$ We have
$19 x=38$
$ \Rightarrow x=2$
Substituting the value of $x$ in equation $(1),$ We have
$5(2)-2 y=2$
$ \Rightarrow 10-2 y=2$
$ \Rightarrow 2 y=10-2$
$ \Rightarrow 2 y=8$
$ \Rightarrow y=\frac{8}{2}$
$ \Rightarrow y=4$
Thus the values of $x$ and $y$ are : $x=2$ and $y=4$.
View full question & answer→Question 164 Marks
Find the values of $m$ and $n$ if :$4^{2 m}=(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2$
Answer$4^{2 \mathrm{\sim m}}=(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2$
$\Rightarrow 4^{2 \mathrm{\sim m}}=(\sqrt{8})^2\ldots .(1)$
and
$(\sqrt[3]{16})^{-\frac{6}{n}}=(\sqrt{8})^2\ldots .(2)$
From$(1)$
$ 4^{2 \mathrm{\sim m}}=(\sqrt{8})^2$
$ \Rightarrow\left(2^2\right)^{2 \mathrm{\sim m}}=\left(\sqrt{2^3}\right)^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=\left[\left(2^3\right)^{\frac{1}{2}}\right]^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=\left[2^{3 \times \frac{1}{2}}\right]^2$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=2^{3 \times \frac{1}{2} \times 2}$
$ \Rightarrow 2^{4 \mathrm{\sim m}}=2^3$
$ \Rightarrow 4 \mathrm{\sim m}=3$
$ \Rightarrow \mathrm{m}=\frac{3}{4}$
From $(2),$ We have
$(3 \sqrt{16})^{-\frac{6}{n}}=(\sqrt{8})^2$
$ \Rightarrow(\sqrt[3]{2 \times 2 \times 2 \times 2})^{-\frac{6}{n}}=(\sqrt{2 \times 2 \times 2})^2$
$ \Rightarrow\left(\sqrt[3]{2^4}\right)^{-\frac{6}{n}}=\left(\sqrt{2^3}\right)^2$
$ \Rightarrow\left[\left(2^4\right)^{\frac{1}{3}}\right]^{-\frac{6}{n}}=\left[\left(2^3\right)^{\frac{1}{2}}\right]^2$
$ \Rightarrow\left[2^{\frac{4}{3}}\right]^{-\frac{6}{n}}=\left[2^{\frac{3}{2}}\right]^2$
$ \Rightarrow 2^{\frac{4}{3} \times\left(-\frac{6}{n}\right)=2^{\frac{3}{2} \times 2}}$
$ \Rightarrow 2^{-\frac{8}{n}}=2^3$
$ \Rightarrow-\frac{8}{\mathrm{n}}=3$
$ \Rightarrow \mathrm{n}=-\frac{8}{3}$
Thus $\mathrm{m}=\frac{3}{4} \mathrm{n}=-\frac{8}{3}$
View full question & answer→Question 174 Marks
Solve : $8 \times 2^{2x} + 4 \times 2^{x + 1} = 1 + 2^x$
Answer$8 \times 2^{2 x}+4 \times 2^{x+1}=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1=1+2^x$
$ \Rightarrow 8 \times\left(2^x\right)^2+4 \times 2^x \times 2^1-1-2^x=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+2^x \times(8-1)-1=0$
$ \Rightarrow 8 \times\left(2^x\right)^2+7\left(2^x\right)-1=0$
$\Rightarrow 8 y^2+7 y-1=0$
$\left[y=2^x\right]$
$ \Rightarrow 8 y^2+8 y-y-1=0$
$ \Rightarrow 8 y(y+1)-1(y+1)=0$
$ \Rightarrow(8 y-1)(y+1)=0$
$ \Rightarrow 8 y=1$ or $y=-1$
$ \Rightarrow y=\frac{1}{8}$ or $y=-1$
$ \Rightarrow 2^x=\frac{1}{8}$ or $2^x=-1$
$ \Rightarrow 2^x=\frac{1}{2^3}$ or $2^x=-1$
$ \Rightarrow 2^x=2^{-3}$ or $2^x=-1$
$ \Rightarrow x=-3$
$[\because 2^x=-1]$ is not possible.
View full question & answer→Question 184 Marks
Simplify :$\frac{3 \times 27^{n+1}+9 \times 3^{3 n-1}}{8 \times 3^{3 n}-5 \times 27^n}$
Answer$ \frac{3 \times 27^{n+1}+9 \times 3^{3 n-1}}{8 \times 3^{3 n}-5 \times 27^n}$
$ =\frac{3 \times(3 \times 3 \times 3)^{n+1}+3 \times 3 \times 3^{3 n-1}}{2 \times 2 \times 2 \times 3^{3 n}-5 \times(3 \times 3 \times 3)^n}$
$ =\frac{3 \times\left(3^3\right)^{n+1}+3^2 \times 3^{3 n-1}}{2^3 \times 3^{3 n}-5 \times\left(3^3\right)^n}$
$ =\frac{3 \times 3^{3 n+3}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n+3+1}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n+4}+3^{3 n+1}}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n} \times 3^4+3^{3 n} \times 3^1}{2^3 \times\left(3^3\right)^n-5 \times\left(3^3\right)^n}$
$ =\frac{3^{3 n}\left(3^4+3^1\right)}{\left(3^3\right)^n(8-5)}$
$ =\frac{3^{3 n}\left(3^4+3^1\right)}{3^{3 n} \times 3}$
$ =\frac{3 \times 3 \times 3 \times 3+3}{3}$
$ =\frac{81+3}{3}$
$ =\frac{84}{3}$
$ =28$
View full question & answer→Question 194 Marks
If $2160 = 2^a. 3^b. 5^c$, find $a, b$ and $c.$ Hence calculate the value of $3^a \times 2^{-b} \times 5^{-c}.$
Answer$2160=2^a \times 3^b \times 5^c$
$ \Rightarrow 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5=2^a \times 3^b \times 5^c$
$ \Rightarrow 2^4 \times 3^3 \times 5^1=2^a \times 3^b \times 5^c$
$ \Rightarrow 2^a \times 3^b \times 5^c=2^4 \times 3^3 \times 5^1$
Comparing powers of $2,3$ and $5$ on the both sides of equation, We have
$a=4 ; b=3$ and $c=1$
Hence,
Value of $3^a \times 2^{-b} \times 5^{-c}$
$=3^4 \times 2^{-3} \times 5^{-1}$
$ =3 \times 3 \times 3 \times 3 \times \frac{1}{2^3} \times \frac{1}{5}$
$=81 \times \frac{1}{2 \times 2 \times 2} \times \frac{1}{5}$
$=81 \times \frac{1}{8} \times \frac{1}{5}$
$=\frac{81}{40}$
$=2 \frac{1}{40}$
View full question & answer→Question 204 Marks
Simplify the following and express with positive index :$\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{-1}$
Answer$ {\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{-1}}$
$ =\frac{1}{\left[1-\left\{1-(1-n)^{-1}\right\}^{-1}\right]^{+} 1}$
$ =\frac{1}{1-\frac{1}{1-(1-n)^{-1}}}$
$ =\frac{1}{1-\frac{1}{1-\frac{1}{1-\pi}}}$
$ =\frac{1}{1-\frac{\frac{1}{1(1-n)-1}}{1-n}}$
$ =\frac{1}{1-\frac{\frac{1}{-n}}{1-n}}$
$ =\frac{1}{1-\frac{1-n}{-n}}$
$ =\frac{1}{1+\frac{1-n}{n}}$
$ =\frac{1}{\frac{n+(1-n)}{n}}$
$ =\frac{\frac{1}{n+1-n}}{n}$
$ =\frac{n}{1}$
$ =\mathrm{n}$
View full question & answer→Question 214 Marks
Evaluate :$\left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$
Answer$ \left(\frac{27}{8}\right)^{\frac{2}{3}}-\left(\frac{1}{4}\right)^{-2}+5^0$
$ =\left(\frac{3 \times 3 \times 3}{2 \times 2 \times 2}\right)^{\frac{2}{3}}-\left(\frac{1 \times 1}{2 \times 2}\right)^{-2}+5^0$
$ =\left[\left(\frac{3}{2}\right)^3\right]^{\frac{2}{3}}-\left[\left(\frac{1}{2}\right)^2\right]^{-2}+1$
$ =\left(\frac{3}{2}\right)^{3 \times \frac{2}{3}}-\left(\frac{1}{2}\right)^{2 \times(-2)}+1$
$ =\left(\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^{-4}+1$
$ =\frac{3}{2} \times \frac{3}{2}-\frac{1}{\left(\frac{1}{2}\right)^4}+1$
$ =\frac{9}{4}-\frac{1}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}}+1$
$ =\frac{9}{4}-\frac{1}{\frac{1}{16}}+1$
$ =\frac{9}{4}-16+1$
$ =\frac{9-64+4}{4}$
$ =\frac{-51}{4}$
View full question & answer→Question 224 Marks
Evaluate :$\left(\frac{16}{81}\right)^{-\frac{3}{4}} \times\left(\frac{49}{9}\right)^{\frac{3}{2}} \div\left(\frac{343}{216}\right)^{\frac{2}{3}}$
Answer$ \left(\frac{16}{81}\right)^{-\frac{3}{4}} \times\left(\frac{49}{9}\right)^{\frac{3}{2}} \div\left(\frac{343}{216}\right)^{\frac{2}{3}}$
$ =\left(\frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3}\right)^{-\frac{3}{4}} \times\left(\frac{7 \times 7}{3 \times 3}\right)^{\frac{3}{2}} \div\left(\frac{7 \times 7 \times 7}{6 \times 6 \times 6}\right)^{\frac{2}{3}}$
$ =\left[\left(\frac{2}{3}\right)^4\right]^{-\frac{3}{4}} \times\left[\left(\frac{7}{3}\right)^2\right]^{\frac{3}{2}} \div\left[\left(\frac{7}{6}\right)^3\right]^{\frac{2}{3}}$
$ =\left(\frac{2}{3}\right)^{4 \times-\frac{3}{4}} \times\left(\frac{7}{3}\right)^{2 \times \frac{3}{2}} \div\left(\frac{7}{6}\right)^{3 \times \frac{2}{3}}$
$ =\left(\frac{2}{3}\right)^{-3} \times\left(\frac{7}{3}\right)^3 \div\left(\frac{7}{6}\right)^2$
$ =\frac{1}{\left(\frac{2}{3}\right)^3} \times\left(\frac{7}{3}\right)^3 \times \frac{1}{\left(\frac{7}{6}\right)^2}$
$ =\frac{1}{\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1}{\frac{7}{6} \times \frac{7}{6}}$
$ =\frac{1 \times 3 \times 3 \times 3}{2 \times 2 \times 2} \times \frac{7}{3} \times \frac{7}{3} \times \frac{7}{3} \times \frac{1 \times 6 \times 6}{7 \times 7}$
$ =\frac{7 \times 3 \times 3}{2}$
$ =\frac{63}{2}$
$ =31.5$
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