Q. 73 - Precalculus Enhanced with Graphing Utilities

Question

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

$3^{x^2-7}={27}^{2x}$

Step-by-Step Solution

Answer

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

$-1,7$

Step 1. Given equation

$3^{x^2-7}={27}^{2x}$

Step 2. Solving exponentially

$$\begin{gathered} 3^{x^2-7}={27}^{2x} \\ {3}^{x^2-7}=3^{3\left(2x\right)} \\ {3}^{x^2-7}=3^{6x} \end{gathered}$$

By exponential property

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

Using this :

$$\begin{gathered} x^2-6x-7=0 \\ \Rightarrow x^2-7x+x-7=0 \\ \Rightarrow x(x-7)+1(x-7)=0 \\ \Rightarrow (x+1)(x-7)=0 \\ \Rightarrow x=-1 or x=7 \end{gathered}$$

Step 3. Verifying Graphically

Sketching the graphs of 

$3^{x^2-7}\&{27}^{2x}$

Where they will intersect will be the solution .

They intersect at -1,7

hence verified