3^x+1=18/3^x+25

Solution for 3^x+1=18/3^x+25 equation:

3^x+1=18/3^x+25

We move all terms to the left:

3^x+1-(18/3^x+25)=0


Domain of the equation: 3^x+25)!=0

x∈R


We get rid of parentheses

3^x-18/3^x-25+1=0

We multiply all the terms by the denominator


3^x*3^x-25*3^x+1*3^x-18=0

Wy multiply elements

9x^2-75x+3x-18=0

We add all the numbers together, and all the variables

9x^2-72x-18=0

a = 9; b = -72; c = -18;
Δ = b2-4ac
Δ = -722-4·9·(-18)
Δ = 5832
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:

$\sqrt{\Delta}=\sqrt{5832}=\sqrt{2916*2}=\sqrt{2916}*\sqrt{2}=54\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-54\sqrt{2}}{2*9}=\frac{72-54\sqrt{2}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+54\sqrt{2}}{2*9}=\frac{72+54\sqrt{2}}{18}
$