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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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Is the log a natural log? Or base 10? –  ncmathsadist Oct 11 '12 at 1:17
    
@ncmathsadist Sorry, I should have specified. Its base 10. –  abc Oct 11 '12 at 1:18
    
Are you looking for real solutions or complex solutions? –  S.B. Oct 11 '12 at 1:19
    
@S.B. I am looking for all real solutions. –  abc Oct 11 '12 at 1:20
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@abc: It can have at most one real solution, because the LHS is strictly increasing while the RHS is strictly decreasing (actually, it has exactly one solution). Obviously you must look at $x>15$. I'm not sure if you can find a solution explicitly, but you can solve it numerically. –  S.B. Oct 11 '12 at 1:24

3 Answers 3

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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Thanks Robert. The solution is approximately 16 (which I found through just substituting the values in). Sorry, but my mathematical skills are relatively elementary - what is a "closed form solution"? The Wikipedia article on it was very rigorous and hard to understand. –  abc Oct 11 '12 at 1:29
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@abc: "closed form" solution means a formula using only elementary operations (addition, substraction, multiplication, division) plus a handful of functions (exponentiation, roots and logarithms, trigonometric). –  Javier Badia Oct 11 '12 at 1:54
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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. –  Robert Israel Oct 11 '12 at 4:59

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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Thanks. What you did above seems to be just an approximation - isn't there a formal way to find an exact solution? –  abc Oct 11 '12 at 1:50
    
Exact solution is too difficult to find. –  minthao_2011 Oct 11 '12 at 2:17
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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. –  coffeemath Oct 11 '12 at 2:25

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. –  TMM Dec 31 '12 at 0:42

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