Question

If ${\left(\sqrt[3]{34}\right)}^{2x+\frac{1}{2}}={\left(\sqrt[3]{38}\right)}^5$, then then value of $x$ is $3\frac{1}{m}$.
So, $m$ is:

Answer 1

${\left(\sqrt[3]{34}\right)}^{2x+\frac{1}{2}}={\left(\sqrt[3]{38}\right)}^5$

$\Rightarrow {\left({\left(2^2\right)}^{\frac{1}{3}}\right)}^{\frac{4x+1}{2}}={\left({\left(2^3\right)}^{\frac{1}{3}}\right)}^5$

$\Rightarrow {\left(2^{2\cdot \frac{1}{3}}\right)}^{\frac{4x+1}{2}}={\left(2^{3\cdot \frac{1}{3}}\right)}^5$

$\Rightarrow \left(2^{\frac{2}{3}\cdot \frac{4x+1}{2}}\right)=2^5$

$\Rightarrow 2^{\frac{4x+1}{3}}=2^5$

Now, we have equation for $x$

$\Rightarrow \frac{4x+1}{3}=5$

$\Rightarrow 4x+1=15$

$\Rightarrow 4x=14$

$\Rightarrow x=\frac{14}{4}$

$\Rightarrow x=3\frac{2}{4}$

$\Rightarrow x=3\frac{1}{2}$