3^(x^2)-42=1/(3^x)

Solution for 3^(x^2)-42=1/(3^x) equation:

3^(x^2)-42=1/(3^x)

We move all terms to the left:

3^(x^2)-42-(1/(3^x))=0


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

x!=0/1

x!=0

x∈R


We get rid of parentheses

3^x^2-1/3^x-42=0

We multiply all the terms by the denominator


3^x^2*3^x-42*3^x-1=0

Wy multiply elements

9x^2-126x-1=0

a = 9; b = -126; c = -1;
Δ = b2-4ac
Δ = -1262-4·9·(-1)
Δ = 15912
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:

$\sqrt{\Delta}=\sqrt{15912}=\sqrt{36*442}=\sqrt{36}*\sqrt{442}=6\sqrt{442}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-126)-6\sqrt{442}}{2*9}=\frac{126-6\sqrt{442}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-126)+6\sqrt{442}}{2*9}=\frac{126+6\sqrt{442}}{18}
$