Chapter 6, Problem 24CR

To determine

To solve the equation:

  ${\left(\frac{1}{16}\right)}^{3x}={64}^{2\left(x+8\right)}$

Answer to Problem 24CR

  $x=-4$

Explanation of Solution

Given:

  ${\left(\frac{1}{16}\right)}^{3x}={64}^{2\left(x+8\right)}$

Concept Used:

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:

  ${\left(\frac{1}{16}\right)}^{3x}={64}^{2\left(x+8\right)}$

First, simplify the equation so the base of the equation is same, as

  $\begin{array}{l}{\left(\frac{1}{16}\right)}^{3x}={64}^{2\left(x+8\right)}\\ \Rightarrow {\left(\frac{1}{4^2}\right)}^{3x}={\left(4^3\right)}^{2\left(x+8\right)}\\ \Rightarrow {\left(4^{-2}\right)}^{3x}={\left(4\right)}^{6\left(x+8\right)}\\ \Rightarrow {\left(4\right)}^{-6x}={\left(4\right)}^{6\left(x+8\right)}\end{array}$

Now, since base of above powers are same; thus

  $\begin{array}{c}{\left(4\right)}^{-6x}={\left(4\right)}^{6\left(x+8\right)}\\ \Rightarrow -6x=6\left(x+8\right)\\ \Rightarrow \frac{-6x}{6}=x+8\\ \Rightarrow -x=x+8\\ \Rightarrow x+x+8=0\\ \Rightarrow 2x+8=0\\ \Rightarrow 2x=-8\\ \Rightarrow x=-4\end{array}$

Thus, the solution of given equation is:

  $x=-4$