3^x-4=45/3^x

Solution for 3^x-4=45/3^x equation:

3^x-4=45/3^x

We move all terms to the left:

3^x-4-(45/3^x)=0


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

x!=0/1

x!=0

x∈R


We get rid of parentheses

3^x-45/3^x-4=0

We multiply all the terms by the denominator


3^x*3^x-4*3^x-45=0

Wy multiply elements

9x^2-12x-45=0

a = 9; b = -12; c = -45;
Δ = b2-4ac
Δ = -122-4·9·(-45)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-42}{2*9}=\frac{-30}{18}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+42}{2*9}=\frac{54}{18}
=3
$