3^-4x-5=(1/27)^2X+10

Solution for 3^-4x-5=(1/27)^2X+10 equation:

3^-4x-5=(1/27)^2x+10

We move all terms to the left:

3^-4x-5-((1/27)^2x+10)=0


Domain of the equation: 27)^2x+10)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-4x-((+1/27)^2x+10)-5+3^=0

We add all the numbers together, and all the variables

-4x-((+1/27)^2x+10)=0

We multiply all the terms by the denominator


-4x*27)^2x+10)-((+1=0

Wy multiply elements

-108x^2+1=0

a = -108; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-108)·1
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*-108}=\frac{0-12\sqrt{3}}{-216}
=-\frac{12\sqrt{3}}{-216}
=-\frac{\sqrt{3}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-108}=\frac{0+12\sqrt{3}}{-216}
=\frac{12\sqrt{3}}{-216}
=\frac{\sqrt{3}}{-18}
$