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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'm not sure why this question was downvoted. It seems like a legitimate question to me. – EuYu Oct 20 '12 at 5:32
Are you familiar with the Lambert-W Function (, hence see:… – Amzoti Oct 20 '12 at 5:37
(I didn't downvote) maybe because there are similar problems already on here? I swear I've seen a few, but I just searched & turned up nothing, so I can't blame you for not finding one! Similar one:… – NeuroFuzzy Oct 20 '12 at 5:39

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 $$x_1= -1.020199952\,, x_2=568.2993002 \,. $$

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Why is this answer not marked as solution? The question is definitely answered by this answer. – pisoir Jan 23 '14 at 17:33
@pisoir: Thanks for the comment. I really appreciate it. – Mhenni Benghorbal Jan 24 '14 at 2:14

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.

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(+1) on the frustration of finding out that the solution of your problem is a Lambert solution :P – Pragabhava Oct 20 '12 at 5:44

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.

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