Question

If $3^{x+8}={27}^{2x+1}$, then the value of
${[{\left(\frac{\sqrt{289}}{\sqrt[3]{3216}}\right)}^x÷{\left(\frac{17}{\sqrt[4]{41296}}\right)}^x]}^{\frac{1}{2}}$ is

• A. 1
• B. 0
• C. $\frac{17}{6}$
• D. $\frac{6}{17}$

Answer 1

The correct option is A 1

$3^{x+8}={27}^{2x+1}$
$\Rightarrow 3^{x+8}=(3^3{)}^{2x+1}$
$\Rightarrow 3^{x+8}=3^{6x+3}$
Base are same so
$x+8=6x+3$
$x-6x=3-8\Rightarrow -5x=-5\Rightarrow x=1$
If x=1 then the value of ${[{\left(\frac{\sqrt{289}}{\sqrt[3]{3216}}\right)}^x÷{\left(\frac{17}{\sqrt[4]{41296}}\right)}^x]}^{\frac{1}{2}}$
$\Rightarrow {[\frac{\sqrt{{\left(17\right)}^2}}{3\sqrt{{\left(6\right)}^3}}÷\frac{17}{4\sqrt{{\left(6\right)}^4}}]}^{\frac{1}{2}}$
$\Rightarrow {[\frac{17}{6}\times \frac{6}{17}]}^{\frac{1}{2}}$
$\Rightarrow {\left[1\right]}^{\frac{1}{2}}\Rightarrow \sqrt{1}=1$