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I've searched for some answer already, but couldn't find any solution to this problem. Apparently, there's no rule for the product of two logarithms. How would I then find the exact solution of this problem? $$ \log(x) = \log(100x) \, \log(2) $$

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How would I then find the exact solution of this problem?

Manipulate the equation to isolate $x$. \begin{align*} \log(x) &= (\log(100)+\log(x))\log(2) \\ \log(x) &=\log(100)\log(2)+\log(x)\log(2)\\ \log(x)-\log(x)\log(2)&=\log(100)\log(2)\\ \log(x)(1-\log(2))&=\log(100)\log(2) \\ \log(x)&=\log(100)\log(2)/(1-\log(2))\\ \end{align*} Then resolve $x$ with whatever base your logarithm is using. E.g. with base 10, $$x\approx7.267$$

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  • $\begingroup$ This should be it if I look at how it's worked out. $\endgroup$ – P.Yntema Oct 16 '16 at 18:43
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$\log(x) = (\log(100) + \log(x))\cdot\log2$

$\log(x) = \log(100)\cdot\log(2) + \log(x)\cdot\log(2)$

This is a linear equation in $\log(x)$!

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  • $\begingroup$ One variable does not a line make, but a point. $\endgroup$ – Ax. Oct 13 '16 at 20:14
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    $\begingroup$ @Ax. That's linear as in "polynomial equation of degree 1", not as in "defining a line". $\endgroup$ – Hong Ooi Oct 14 '16 at 7:45
  • $\begingroup$ Oh right, the other variable is the base of the logarithm. $\endgroup$ – Ax. Oct 14 '16 at 13:17
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    $\begingroup$ @Ax. Set the $x$ axis to be $\log(x)$ and the $y$ axis to be regular $y$. Then it becomes a line. $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 13:16
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    $\begingroup$ @Ax. this may help clarify things en.wikipedia.org/wiki/Linear_equation $\endgroup$ – Giuseppe Oct 17 '16 at 13:24
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The question does not specify the base $B$ of the logarithm, but it will affect the solution, so we make it explicit: \begin{align} \log_B(x) &= \log_B(100\, x) \, \log_B(2) \\ &= (\log_B(100) + \log_B(x)) \, \log_B(2) \iff \\ (1 - \log_B(2)) \log_B(x) &= \log_B(2) \log_B(100) \\ \end{align} For $B = 2$ the LHS vanishes and we have no solution, as the logarithms on the RHS do not vanish.

For $B \ne 2$ we can continue: \begin{align} \log_B(x) = \frac{\log_B(2) \, \log_B(100)}{1 - \log_B(2)} = f(B) \iff \\ x = B^{f(B)} = B^{(\log_B(2) \, \log_B(100))/(1 - \log_B(2))} \end{align}

For $B=e$ one gets $$ x = e^{f(e)} = e^{10.4025\dotsb} = 32944.48\dotsb $$

Here are graphs of $f(B)$:

close view farer view

(Links to larger versions: left, right)

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    $\begingroup$ I believe that is the incorrect approximate value of the solution. $\endgroup$ – Ian Limarta Oct 13 '16 at 15:48
  • $\begingroup$ I see now that you used value e as your base. The solution, however, is still wrong. $\endgroup$ – Ian Limarta Oct 13 '16 at 15:53
  • $\begingroup$ Ah yes. I separated it now. $\endgroup$ – mvw Oct 13 '16 at 17:23
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    $\begingroup$ If B is not specified, it automatically means that it is 10. $\endgroup$ – P.Yntema Oct 16 '16 at 18:44
  • $\begingroup$ While true, people do use e to denote log. Personally, I find the log notation of base 10 to be silly. But hey, base 10 is "silent" wherever you go. $\endgroup$ – Ian Limarta Oct 20 '16 at 19:30
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Fill in details after you check the basic properties of logarithms, and assuming $\;\log=\log_{10}\;$:

$$\log x=\log 100x\cdot\log2=\left(\log100+\log x\right)\log2\implies$$

$$(1-\log2)\log x=2\log2\implies\ldots$$

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  • $\begingroup$ A term has disappeared $\endgroup$ – egreg Oct 13 '16 at 15:25
  • $\begingroup$ @egreg I don't think so....which one? $\endgroup$ – DonAntonio Oct 13 '16 at 15:58
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    $\begingroup$ @egreg My logic was that such an elementary question surely is of high school level, and there they usually use $\;\log=\log_{10}\;$, as shown in calculators. That's the only reason. $\endgroup$ – DonAntonio Oct 13 '16 at 16:25
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    $\begingroup$ @CarlWitthoft My "standard" notation in high school was ln for natural logs, and log for base 10 logs. $\endgroup$ – Brian J Oct 13 '16 at 20:08
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    $\begingroup$ @CarlWitthoft I have never, ever heard of that "standard notation", by which I mean some widely used and more or less international notation. In my case, and my children's, the notation was as Brian says. $\endgroup$ – DonAntonio Oct 13 '16 at 20:40

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