4^x+3=1/2^x

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

4^x+3=1/2^x

We move all terms to the left:

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


Domain of the equation: 2^x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

8x^2+6x-1=0

a = 8; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·8·(-1)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{17}}{2*8}=\frac{-6-2\sqrt{17}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{17}}{2*8}=\frac{-6+2\sqrt{17}}{16}
$