Q. 72 - Precalculus Enhanced with Graphing Utilities

Question

solve each equation. Verify your results using a graphing utility.

$9^{-x+15}={27}^x$

Step-by-Step Solution

Answer

The value of x for which the above equation is valid is :

$6$

Step 1. Given equation

$9^{-x+15}={27}^x$

Step 2. Solving exponentially

$$\begin{gathered} 9^{-x+15}={27}^x \\ \Rightarrow 3^{2(-x+15)}=3^{3\left(x\right)} \\ \Rightarrow 3^{(-2x+30)}=3^{3x} \end{gathered}$$

By exponential property 

$$\begin{gathered} a^v=a^u \\ \Rightarrow v=u \end{gathered}$$

Using this :

Step 3. Verifying Graphically

Sketching the graphs of 

$9^{-x+15} \& {27}^x$

Where they will intersect will be the solution .

they intersect at point 6

Hence verified