(7/3x)+1=2^x

Solution for (7/3x)+1=2^x equation:

(7/3x)+1=2^x

We move all terms to the left:

(7/3x)+1-(2^x)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/3x)-2^x+1=0

We get rid of parentheses

7/3x-2^x+1=0

We multiply all the terms by the denominator


-2^x*3x+1*3x+7=0

Wy multiply elements

-6x^2+3x+7=0

a = -6; b = 3; c = +7;
Δ = b2-4ac
Δ = 32-4·(-6)·7
Δ = 177
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{-(3)-\sqrt{177}}{2*-6}=\frac{-3-\sqrt{177}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{177}}{2*-6}=\frac{-3+\sqrt{177}}{-12}
$