2^x-1=1/8^x

Solution for 2^x-1=1/8^x equation:

2^x-1=1/8^x

We move all terms to the left:

2^x-1-(1/8^x)=0


Domain of the equation: 8^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

2^x-1/8^x-1=0

We multiply all the terms by the denominator


2^x*8^x-1*8^x-1=0

Wy multiply elements

16x^2-8x-1=0

a = 16; b = -8; c = -1;
Δ = b2-4ac
Δ = -82-4·16·(-1)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*16}=\frac{8-8\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*16}=\frac{8+8\sqrt{2}}{32}
$