6^-3x-4=(1/6)^6x

Solution for 6^-3x-4=(1/6)^6x equation:

6^-3x-4=(1/6)^6x

We move all terms to the left:

6^-3x-4-((1/6)^6x)=0


Domain of the equation: 6)^6x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-3x-((+1/6)^6x)-4+6^=0

We add all the numbers together, and all the variables

-3x-((+1/6)^6x)=0

We multiply all the terms by the denominator


-3x*6)^6x)-((+1=0

Wy multiply elements

-18x^2+1=0

a = -18; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-18)·1
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-18}=\frac{0-6\sqrt{2}}{-36}
=-\frac{6\sqrt{2}}{-36}
=-\frac{\sqrt{2}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-18}=\frac{0+6\sqrt{2}}{-36}
=\frac{6\sqrt{2}}{-36}
=\frac{\sqrt{2}}{-6}
$