Math Doubts

Calculate the values of $x$ and $y$, if $z^3 = \sqrt{36^5}$, $y = x^{-5}$ and $x = \log_{6} (z) -\dfrac{2}{3}$

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

$(1) \,\,\,\,\,\,$ $z^3 = \sqrt{36^5}$

$(2) \,\,\,\,\,\,$ $x = \log_{6} (z) -\dfrac{2}{3}$

$(2) \,\,\,\,\,\,$ $y = x^{-5}$

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 $z$.

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

$\implies z^3 = \sqrt{{(6^2)}^5}$

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

$\implies z^3 = \sqrt{6^{2 \times 5}}$

$\implies z^3 = \sqrt{6^{5 \times 2}}$

$\implies z^3 = \sqrt{{(6^5)}^2}$

$\implies z^3 = 6^5$

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

$\implies \sqrt[\displaystyle 3]{z^3} = \sqrt[\displaystyle 3]{6^5}$

$\therefore \,\,\,\,\,\, z = {(6)}^{\dfrac{5}{3}}$

Step: 2

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

$x = \log_{6} (z) -\dfrac{2}{3}$

$\implies x = \log_{6} {(6)}^{\dfrac{5}{3}} -\dfrac{2}{3}$

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

$\implies x = \dfrac{5}{3} \log_{6} 6 -\dfrac{2}{3}$

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

$\implies x = \dfrac{5}{3} \times 1 -\dfrac{2}{3}$

$\implies x = \dfrac{5}{3}-\dfrac{2}{3}$

$\implies x = \dfrac{5-2}{3}$

$\implies x = \dfrac{3}{3}$

$\therefore \,\,\,\,\,\, x = 1$

Step: 3

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

$y = x^{-5}$

$\implies y = {(1)}^{-5}$

$\therefore \,\,\,\,\,\, y = 1$

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



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