# Solving equation with exponentials and addition

Apologies if this question has already been answered elsewhere, but I could not find the answer explicitly (this answer appears related - logarithm of a sum or addition)

How would you solve an equation similar to this:

$2^x + 2 = 2^{2x}$

By inspection, the solution is $x=1$, but I am interested in how a solution could be obtained with different values in the equation. For example, how would you solve:

$2^x + 3 = 2^{2x}$ or $2^x + 2 = 2^{3x}$

I already know that a solution can be found using iterative or numeric methods; I am looking for an algebraic or analytic solution to this kind of equation (or an explanation of why none exist).

EDIT:

Ideally a method for solving the above examples would be applicable to a case such as $2^x + 2 = 3^{x}$.

HINT

The standard method is to consider $t=2^x$ then

$$2^x + 2 = 2^{2x} \implies t^2-t-2=0$$

then solve for $t$ by quadratic equation and for

$$2^x + 2 = 2^{3x} \implies t^3-t-2=0$$

we need to solve first a cubic equation.

Once we have valid solutions for $t>0$ we can obtain $x=\log_2 t$.

• Does this mean that cases where the greatest power of t > 4, there is not always an algebraic solution to the equation? – random.george 11 May 22 '18 at 17:02