3^x+1/3^x=10

Solution for 3^x+1/3^x=10 equation:

3^x+1/3^x=10

We move all terms to the left:

3^x+1/3^x-(10)=0


Domain of the equation: 3^x!=0

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


3^x*3^x-10*3^x+1=0

Wy multiply elements

9x^2-30x+1=0

a = 9; b = -30; c = +1;
Δ = b2-4ac
Δ = -302-4·9·1
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-12\sqrt{6}}{2*9}=\frac{30-12\sqrt{6}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+12\sqrt{6}}{2*9}=\frac{30+12\sqrt{6}}{18}
$