How to solve a + 2^a = 6?

Question

I tried solving the above equation, but was unable to solve it.

How do you basically isolate the exponent and retain the other variable in the same degree.

How to solve a + 2^a = 6?

Answer 1

There is no analytical method to solve this transcendental equation, because it mixes a variable and an exponential. But you can observe that the $LHS$ is a monotonous function (as it is the sum of two growing functions), and there is at most one solution.

By inspection, $2+2^2=6$ and you are done.

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The equation was "arranged" to have an integer solution. Had it not been, you would have had to resort to a numerical method. By trying a few "random" values,

$$0+2^0<6$$ and $$3+2^3>6$$ shows you that the solution lies in $(0,3)$, and you can try intermediate values to get a tighter bracketing. Specific methods exist to reduce the bracketing as fast as possible.

Answer 2

The function $f(x)=x+2^x$ is increasing as $f'(x)=1+\ln{(2)} 2^x \gt 0$ for all $x\in \mathbb{R}$. So, the only solution to the equation given in the set $\mathbb{R}$ is $a=2$.