4^x=2/3x+5

Solution for 4^x=2/3x+5 equation:

4^x=2/3x+5

We move all terms to the left:

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


Domain of the equation: 3x+5)!=0

x∈R


We get rid of parentheses

4^x-2/3x-5=0

We multiply all the terms by the denominator


4^x*3x-5*3x-2=0

Wy multiply elements

12x^2-15x-2=0

a = 12; b = -15; c = -2;
Δ = b2-4ac
Δ = -152-4·12·(-2)
Δ = 321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{321}}{2*12}=\frac{15-\sqrt{321}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{321}}{2*12}=\frac{15+\sqrt{321}}{24}
$