4^x=3/2x+4

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

4^x=3/2x+4

We move all terms to the left:

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


Domain of the equation: 2x+4)!=0

x∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

8x^2-8x-3=0

a = 8; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·8·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{10}}{2*8}=\frac{8-4\sqrt{10}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*8}=\frac{8+4\sqrt{10}}{16}
$