# Apparently cannot be solved using logarithms

This equation clearly cannot be solved using logarithms.

$$3 + x = 2 (1.01^x)$$

Now it can be solved using a graphing calculator or a computer and the answer is $x = -1.0202$ and $x=568.2993$.

But is there any way to solve it algebraically/algorithmically?

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I have solved a question similar to this before. In general, you can have a solution of the equation

$$a^x=bx+c$$

in terms of the Lambert W-function

$$-\frac{1}{\ln(a)}W_k \left( -\frac{1}{b}\ln(a) {{\rm e}^{-{\frac {c\ln(a) }{b}}}} \right)-{\frac {c}{b}} \,.$$

Substituting $a=1.01 \,,b=\frac{1}{2}\,,c=\frac{3}{2}$ and considering the values $k=0$ and $k=-1$, we get the zeroes $$x_1= -1.020199952\,, x_2=568.2993002 \,.$$

Polynomials don’t play nice with exponentials, so no. If you work hard, you might find an answer in terms of the Lambert W function, but if I did I wouldn’t feel much more enlightened.

A standard root finding procedure (such as Newton’s method) should solve the problem for you. You might also be interested in the Lambert W function, which will give you a “closed form” solution, assuming you have access to that function of course.