Chapter 6, Problem 25CR

To determine

To solve the equation:

  ${27}^{2x+2}={81}^{x+4}$

Answer to Problem 25CR

  $x=5$

Explanation of Solution

Given:

  ${27}^{2x+2}={81}^{x+4}$

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:

  ${27}^{2x+2}={81}^{x+4}$

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

  $\begin{array}{l}{27}^{2x+2}={81}^{x+4}\\ \Rightarrow {\left(3^3\right)}^{2x+2}={\left(3^4\right)}^{x+4}\\ \Rightarrow {\left(3\right)}^{3\times 2x+2}={\left(3\right)}^{4\times x+4}\\ \Rightarrow 3^{6x+6}=3^{4x+16}\end{array}$

Now, since base of above powers are same; thus

  $\begin{array}{c}{3}^{6x+6}=3^{4x+16}\\ \Rightarrow 6x+6=4x+16\\ \Rightarrow 6x-4x=16-6\\ \Rightarrow 2x=10\\ \Rightarrow x=\frac{10}{2}\\ \Rightarrow x=5\end{array}$

Thus, the solution of given equation is:

  $x=5$