# How do you solve 4^x + 6(4^-x) = 5?

May 3, 2015

In this way:

Multiply both the members of the equation for ${4}^{x}$, that is always a positive (not zero) term.

${4}^{x} \cdot {4}^{x} + 6 \cdot {4}^{-} x {4}^{x} - 5 \cdot {4}^{x} = 0 \Rightarrow {4}^{2 x} - {5}^{\cdot} {4}^{x} + 6 = 0$.

This is a quadratric equation in the variable ${4}^{x}$, so:

$\Delta = {b}^{2} - 4 a c = 25 - 24 = 1 = {1}^{2}$,

${4}^{x} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{5 \pm 1}{2} \Rightarrow$

${4}^{x} = \left(\frac{5 + 1}{2}\right) = 3 \Rightarrow \ln {4}^{x} = \ln 3 \Rightarrow x \ln 4 = \ln 3 \Rightarrow x = \ln \frac{3}{\ln} 4$;

${4}^{x} = \frac{5 - 1}{2} = 2 \Rightarrow {2}^{2 x} = {2}^{1} \Rightarrow 2 x = 1 \Rightarrow x = \frac{1}{2}$.