5/7^x=5/4^x+4

Solution for 5/7^x=5/4^x+4 equation:

5/7^x=5/4^x+4

We move all terms to the left:

5/7^x-(5/4^x+4)=0


Domain of the equation: 7^x!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 4^x+4)!=0

x∈R


We get rid of parentheses

5/7^x-5/4^x-4=0

We calculate fractions


20x/28x^2+(-35x)/28x^2-4=0

We multiply all the terms by the denominator


20x+(-35x)-4*28x^2=0

Wy multiply elements

-112x^2+20x+(-35x)=0

We get rid of parentheses

-112x^2+20x-35x=0

We add all the numbers together, and all the variables

-112x^2-15x=0

a = -112; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·(-112)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*-112}=\frac{0}{-224}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*-112}=\frac{30}{-224}
=-15/112
$