Chapter 6.4, Problem 58E

To determine

To solve the equation:

  ${\left(\sqrt[5]{3}\right)}^{5x-10}={\left(\sqrt[8]{3}\right)}^{4x}$

Answer to Problem 58E

  $x=4$

Explanation of Solution

Given:

  ${\left(\sqrt[5]{3}\right)}^{5x-10}={\left(\sqrt[8]{3}\right)}^{4x}$

Concept Used:

• $\sqrt[n]{x}=x^{\frac{1}{n}};n\ne 0$
• Two powers with the same positive base(not equals to 1), are equal if and only if their exponents are equal, that is:

  $b^x=b^y\iff x=y;\text{ where }b>0,\text{ and }b\ne 1$

Calculation:

To solve the equation:

  ${\left(\sqrt[5]{3}\right)}^{5x-10}={\left(\sqrt[8]{3}\right)}^{4x}$

First, simplify the equation using property $\sqrt[n]{x}=x^{\frac{1}{n}}$ , as

  $\begin{array}{c}{\left(\sqrt[5]{3}\right)}^{5x-10}={\left(\sqrt[8]{3}\right)}^{4x}\\ \Rightarrow {\left(3^{\frac{1}{5}}\right)}^{5x-10}={\left(3^{\frac{1}{8}}\right)}^{4x}\\ \Rightarrow 3^{\frac{1}{5}\left(5x-10\right)}=3^{\frac{1}{8}\times 4x}\\ \Rightarrow 3^{\frac{1}{5}\times 5x-\frac{1}{5}\times 10}=3^{\frac{1}{2}x}\\ \Rightarrow 3^{x-2}=3^{\frac{1}{2}x}\end{array}$

Now, since base of above powers are same.

Thus

  $\begin{array}{c}{3}^{x-2}=3^{\frac{1}{2}x}\\ \Rightarrow x-2=\frac{x}{2}\\ \Rightarrow x-\frac{x}{2}=2\\ \Rightarrow \frac{2x-x}{2}=2\\ \Rightarrow \frac{x}{2}=2\\ \Rightarrow x=2\times 2\\ \Rightarrow x=4\end{array}$

Thus, the solution of given equation is:

  $x=4$