8/9^x=7^1-x

Solution for 8/9^x=7^1-x equation:

8/9^x=7^1-x

We move all terms to the left:

8/9^x-(7^1-x)=0


Domain of the equation: 9^x!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

8/9^x+x-7^1=0

We multiply all the terms by the denominator


x*9^x-7^1*9^x+8=0

Wy multiply elements

9x^2-63x^2+8=0

We add all the numbers together, and all the variables

-54x^2+8=0

a = -54; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-54)·8
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*-54}=\frac{0-24\sqrt{3}}{-108}
=-\frac{24\sqrt{3}}{-108}
=-\frac{2\sqrt{3}}{-9}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*-54}=\frac{0+24\sqrt{3}}{-108}
=\frac{24\sqrt{3}}{-108}
=\frac{2\sqrt{3}}{-9}
$