Question

Find the values of x in each of the following:

$\left(i\right)2^{5x}÷2^x=\sqrt[5]{52^{20}}$

$\left(ii\right)(2^3{)}^4=(2^2{)}^x$

$\left(iii\right){\left(\frac{3}{5}\right)}^x{\left(\frac{5}{3}\right)}^{2x}=\frac{125}{27}$

$\left(iv\right)5^{x-2}\times 3^{2x-3}=135$

$\left(v\right)2^{x-7}\times 5^{x-4}=1250$

$\left(vi\right)(\sqrt[3]{34}{)}^{2x+\frac{1}{2}}=\frac{1}{32}$

$\left(vii\right)5^{2x+3}=1$

$\left(viii\right)(13{)}^{\sqrt{x}}=4^4-3^4-6$

$\left(ix\right){\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}$

Answer 1

$\left(i\right)2^{5x}÷2^x=\sqrt[5]{52^{20}}\Rightarrow \frac{2^{5x}}{2^x}=(2^{20}{)}^{\frac{1}{5}}\Rightarrow 2^{5x-x}=2^{\frac{20}{5}}\Rightarrow 2^{4x}=2^{\frac{20}{5}}Comparing,weget4x=\frac{20}{5}\Rightarrow x=\frac{20}{5\times 4}=1\therefore x=1$

$\left(ii\right)(2^3{)}^4=(2^2{)}^x\Rightarrow 2^{3\times 4}=2^{2\times x}\Rightarrow 2^{12}=2^{2x}Comparing,weget2x=12\Rightarrow x=\frac{12}{2}=6\therefore x=6$

$\left(iii\right){\left(\frac{3}{5}\right)}^x{\left(\frac{5}{3}\right)}^{2x}=\frac{125}{27}\Rightarrow {\left(\frac{5}{3}\right)}^{-x}{\left(\frac{5}{3}\right)}^{2x}=\frac{5^3}{3^3}\Rightarrow {\left(\frac{5}{3}\right)}^{-x+2x}={\left(\frac{5}{3}\right)}^{3}Comparing,weget-x+2x=3\Rightarrow x=3\therefore x=3$

$\left(iv\right)5^{x-2}\times 3^{2x-3}=1355^{x-2}\times 3^{2x-3}=1355^x{.5}^{-2}{.3}^{2x}{.3}^{-3}=135\frac{5^x{.3}^{2x}}{5^2\times 3^3}=1355^x{.3}^{2x}=135\times 5^2\times 3^3{5}^x\times 3^x\times 3^x=135\times 25\times 27(5\times 3\times 3{)}^x=3\times 3\times 3\times 5\times 5\times 5\times 3\times 3\times 3(45{)}^x=(3\times 5\times 3{)}^3=(45{)}^{3}Comparing,wegetx=3$

$\left(v\right)2^{x-7}\times 5^{x-4}=12502^{x-7}\times 5^{x-4}=1250\Rightarrow 2^x{.2}^{-7}{.5}^x{.5}^{-4}=2\times 5\times 5\times 5\times 5\Rightarrow \frac{2^x{.5}^x}{2^7\times 5^4}=2\times 5\times 5\times 5\times 5\Rightarrow (10{)}^x=2\times 5\times 5\times 5\times 5\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 5\times 5\times 5\times 5\Rightarrow (10{)}^x=2^8\times 5^8=(10{)}^{8}Comparing,wegetx=8$

$\left(vi\right)(\sqrt[3]{34}{)}^{2x+\frac{1}{2}}=\frac{1}{32}\Rightarrow {\left[\right(2^2{)}^{\frac{1}{3}}]}^{2x+\frac{1}{2}}=\frac{1}{2^5}\Rightarrow 2^{\frac{2}{3}(2x+\frac{1}{2})}=2^{-5}\Rightarrow \frac{2}{3}(2x+\frac{1}{2})=-5\Rightarrow \frac{4}{3}x+\frac{1}{3}=-5\Rightarrow 8x+2=-5\times 6\Rightarrow 8x=-30-2\Rightarrow x=\frac{-32}{8}=-4$

$\left(vii\right)5^{2x+3}=1=5^0(\because a^0=1)\therefore 2x+3=0\Rightarrow 2x=-3\therefore x=\frac{-3}{2}$

$\left(viii\right)(13{)}^{\sqrt{x}}=4^4-3^4-6=256-81-6=169=(13{)}^2\therefore \sqrt{x}=2Squaring,x=4$

$\left(ix\right){\left(\sqrt{\frac{3}{5}}\right)}^{x+1}=\frac{125}{27}\Rightarrow {\left(\frac{3}{5}\right)}^{\frac{x+1}{2}}={\left(\frac{3}{5}\right)}^{-3}\left\{\begin{array}{c}\because a^m=\frac{1}{a^{-m}}\end{array}\right\}\frac{x+1}{2}=-3\Rightarrow x+1=-6\Rightarrow x=-7\therefore x=-7$