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I have a formula as follows:

$0.55508389365 = \ln(20 + x)/\ln(240 +x)$

I happen to know that x = 1 (approximately) is the solution. But I'm not sure how I would go about solving this if I didn't know the answer.

By messing around with log manipulation I eventually got to

$240 + x = (20 + x)$ to the power of $1/0.55508389365$

But I'm not sure how you would solve this either!

Is it possible to solve an equation like this? If so how do I do it?

Thanks in advance.

haqnatural
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Martin
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    Where did that ugly decimal number $0.55508389365$ come from? – Mike Pierce Jul 01 '16 at 16:06
  • If I work the equation the other way. So ln(21)/ln(241) – Martin Jul 01 '16 at 16:07
  • Apply Bisection method. You will get a pretty close answer I think – Qwerty Jul 01 '16 at 16:09
  • You might try rewriting $\ln(20+x)=\ln(20)+\ln\left(1+\left(\frac{x}{20}\right)\right)$ and doing the same for $\ln(240+x)$ and use the second or third degree Taylor approximation. For example $\ln\left(1+\left(\frac{x}{20}\right)\right)\approx \left(\frac{x}{20}\right)-\left(\frac{x}{20}\right)^2$. Second order would give a quadratic in $x$ which might give an approximate solution for $x$. Or you could go the cubic approximation which would be much messier. – John Wayland Bales Jul 01 '16 at 20:25
  • p.s. My TI 84 gives the intersection of the two curves $(1, 0.55508389)$. Not everything must have an analytical solution. – John Wayland Bales Jul 01 '16 at 20:33

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