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$3\log_{10}(x-15) = \left(\frac{1}{4}\right)^x$

I am completely lost on how to proceed. Could someone explain how to find any real solution to the above equation?

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    http://www.wolframalpha.com/input/?i=solve%283*log_10%28x-15%29%3D2%5E%28-2x%29%2Cx%292012-10-11

3 Answers 3

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Hint: Consider left and right sides at $x=16$ and $x=17$. You won't find an explicit "closed-form" solution, but you can prove that it exists.

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    "Closed-form" is a rather subjective term: basically you allow "well-known" functions, but people will differ on precisely which ones to include.2012-10-11
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Put \begin{equation*} f(x) = 3\log_{10}(x - 15) - \left(\dfrac{1}{4}\right)^x. \end{equation*} We have $f$ is a increasing function on $(15, +\infty)$. Another way, $f(16)>0 $ and $f(17)>0$. Therefore the given equation has only solution belongs to $(16,17)$.

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    At x=16 we have $log_{10}(16-15)=0$ so f(16)<0. I guess that's what you should have written in your answer anyway, since the statement f(17)>0 is right. You want a sign change to guarantee a zero in between.2012-10-11
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Let $x=16+y$. After approximating $\log(1+y)$ with $y - \frac{y^2}{2}$, $(\frac{1}{4})^y = \exp(-\log(4) y)$ with $1 - \log(4) y$, $\frac{1}{1+\epsilon}$ with $1 - \epsilon$, get $y \approx \frac{\log(10)}{3 \cdot 4^{16}}.$

WIMS Function Calculator gives the exact solution, $1.7870412306 \cdot 10^{-10}$ compared to the approximation, $1.7870412309 \cdot 10^{-10}$.

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    Please check if my LaTeX-editing did not mess up your equations. In any case, (+1) for the nice approach.2012-12-31