How do you solve: $4^{x^2-x-6} = 7$

Question

I know for a fact that it could be solved using logarithms. But i was stuck at a point in the middle of solving and i need help. I would be grateful if you could show how to find the value/s of x ( logarithms method )

Answer 1

This is the process you would use

$4^{x^2-x-6}=7 \longrightarrow \ln(4^{x^2-x-6})=\ln{7} \longrightarrow \ln{4} (x^2-x-6) =\ln{7} \longrightarrow x^2-x-6 = \frac{\ln{7}}{\ln{4}} \longrightarrow x^2-x-(6+\frac{\ln{7}}{\ln{4}})=0$

Now you could use the quadratic equation.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{1\pm\sqrt{1-4*(-6-\frac{\ln{7}}{\ln{4}})}}{2}$