1/2^x+4=2/3^x

Solution for 1/2^x+4=2/3^x equation:

1/2^x+4=2/3^x

We move all terms to the left:

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


Domain of the equation: 2^x!=0

x!=0/1

x!=0

x∈R



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

x!=0/1

x!=0

x∈R


We get rid of parentheses

1/2^x-2/3^x+4=0

We calculate fractions


3x/6x^2+(-4x)/6x^2+4=0

We multiply all the terms by the denominator


3x+(-4x)+4*6x^2=0

Wy multiply elements

24x^2+3x+(-4x)=0

We get rid of parentheses

24x^2+3x-4x=0

We add all the numbers together, and all the variables

24x^2-1x=0

a = 24; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·24·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*24}=\frac{0}{48}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*24}=\frac{2}{48}
=1/24
$