16^x+1=1/4^x

Solution for 16^x+1=1/4^x equation:

16^x+1=1/4^x

We move all terms to the left:

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


Domain of the equation: 4^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

16^x-1/4^x+1=0

We multiply all the terms by the denominator


16^x*4^x+1*4^x-1=0

Wy multiply elements

64x^2+4x-1=0

a = 64; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·64·(-1)
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{17}}{2*64}=\frac{-4-4\sqrt{17}}{128}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{17}}{2*64}=\frac{-4+4\sqrt{17}}{128}
$