I am not sure, but it's an equation...

Question

$$2x + 2^x = 4$$ So it's clear that $x$ will equal $1$ but how can it be solved through an algebraic method to be able to determine $x$ with more complex numbers.

I've tried to transform the exponent into a logarithm

$ 2x + \log_2 2^x = 4$

$2x +x = 4$

$x = \frac43$

But it didn't work...

Answer 1

If you consider the function $$ f(x)=2x+2^x $$ you see that $$ \lim_{x\to-\infty}f(x)=-\infty, \qquad \lim_{x\to\infty}f(x)=\infty $$ and $$ f'(x)=2+2^x\log 2>0 $$ for every $x$. Thus the function is strictly increasing and so it takes exactly once every possible real value.

Since, clearly, $$ f(1)=2\cdot1+2^1=4 $$ you see that the only solution of your equation is $x=1$.

There is no simple “algebraic” solution. What you can say is that, for every real $a$, there is a unique $x$ such that $2x+2^x=a$. Such $x$ can be determined with the desired accuracy with some approximation method.

Answer 2

As you said, $x=1$ is solution. Moreover, if $f(x)=2x+2^x$,$$f'(x)=2+\ln(2)2^x>0$$ and thus, $f$ is injective. Therefore, $x=1$ is the unique solution.