36^-3x+3=(1/216)^x+1

Solution for 36^-3x+3=(1/216)^x+1 equation:

36^-3x+3=(1/216)^x+1

We move all terms to the left:

36^-3x+3-((1/216)^x+1)=0


Domain of the equation: 216)^x+1)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/216)^x+1)+3+36^=0

We add all the numbers together, and all the variables

-3x-((+1/216)^x+1)=0

We multiply all the terms by the denominator


-3x*216)^x+1)-((+1=0

Wy multiply elements

-648x^2+1=0

a = -648; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-648)·1
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*-648}=\frac{0-36\sqrt{2}}{-1296}
=-\frac{36\sqrt{2}}{-1296}
=-\frac{\sqrt{2}}{-36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*-648}=\frac{0+36\sqrt{2}}{-1296}
=\frac{36\sqrt{2}}{-1296}
=\frac{\sqrt{2}}{-36}
$