Calculate the values of x and y

Calculate the values of $x$ and $y$, if $z^3=\sqrt{{36}^5}$, $y=x^{-5}$ and $x={\log }_6\left(z\right)-\frac{2}{3}$

Three mathematical statements are given to find the values of $x$ and $y$ in this logarithm problem.

$\left(1\right)$ $z^3=\sqrt{{36}^5}$

$\left(2\right)$ $x={\log }_6\left(z\right)-\frac{2}{3}$

$\left(2\right)$ $y=x^{-5}$

The first statement expresses the value of $z$ in terms of a radical having an exponential term as its radicand. The second statement is useful to find the value of $x$ by substituting the value of $z$ in it. Finally, the third statement is useful to find the value $y$ by substituting the value of $x$ in it.

Step: 1

Solve the first statement and find the value of $z$.

$z^3=\sqrt{{36}^5}$

$⟹z^3=\sqrt{{\left(6^2\right)}^5}$

Apply the power rule of an exponential term to simplify this expression.

$⟹z^3=\sqrt{6^{2\times 5}}$

$⟹z^3=\sqrt{6^{5\times 2}}$

$⟹z^3=\sqrt{{\left(6^5\right)}^2}$

$⟹z^3=6^5$

Take cube root both sides to find the value of $z$.

$⟹\sqrt[3]{3z^3}=\sqrt[3]{36^5}$

$\therefore z={\left(6\right)}^{\frac{5}{3}}$

Step: 2

Now, substitute the value of $z$ in the second statement to obtain the value of $x$.

$x={\log }_6\left(z\right)-\frac{2}{3}$

$⟹x={\log }_6{\left(6\right)}^{\frac{5}{3}}-\frac{2}{3}$

Use the power law of logarithm of an exponential term to simplify the equation.

$⟹x=\frac{5}{3}{\log }_{6}6-\frac{2}{3}$

According to logarithm of base rule, the logarithm of a number to same number is one.

$⟹x=\frac{5}{3}\times 1-\frac{2}{3}$

$⟹x=\frac{5}{3}-\frac{2}{3}$

$⟹x=\frac{5-2}{3}$

$⟹x=\frac{3}{3}$

$\therefore x=1$

Step: 3

Now substitute the value of $x$ in third algebraic equation to get the value of $y$.

$y=x^{-5}$

$⟹y={\left(1\right)}^{-5}$

$\therefore y=1$

Therefore, it is derived that value of $x$ is equal to the value of $y$ and it is $1$. It is written as $x=y=1$