4^(2x)=1/16

Solution for 4^(2x)=1/16 equation:

4^(2x)=1/16

We move all terms to the left:

4^(2x)-(1/16)=0

We add all the numbers together, and all the variables

4^2x-(+1/16)=0

We get rid of parentheses

4^2x-1/16=0

We multiply all the terms by the denominator


4^2x*16-1=0

Wy multiply elements

64x^2-1=0

a = 64; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·64·(-1)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*64}=\frac{-16}{128}
=-1/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*64}=\frac{16}{128}
=1/8
$