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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?

  • 0
    (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: http://math.stackexchange.com/questions/61774/how-do-i-solve-this-equation-involving-a-logarithm2012-10-20

3 Answers 3

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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 \,. $

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    @pisoir: Thanks for the comment. I really appreciate it.2014-01-24
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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 proble$m$ is a Lambert solutio$n$ :P2012-10-20
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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.