Multiplying two logarithms

Question

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? $$ \log(x) = \log(100x) \, \log(2) $$

score 19 · Answer 1 · answered Oct 13 '16 at 15:08

19

$\log(x) = (\log(100) + \log(x))\cdot\log2$

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

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

answered Oct 13 '16 at 15:08

Giuseppe

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One variable does not a line make, but a point. – Ax. Oct 13 '16 at 20:14
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@Ax. That's linear as in "polynomial equation of degree 1", not as in "defining a line". – Hong Ooi Oct 14 '16 at 07:45
Oh right, the other variable is the base of the logarithm. – Ax. Oct 14 '16 at 13:17
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@Ax. Set the $x$ axis to be $\log(x)$ and the $y$ axis to be regular $y$. Then it becomes a line. – Simply Beautiful Art Oct 15 '16 at 13:16
@SimpleArt, where is $y$? – Ax. Oct 17 '16 at 13:15
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@Ax. this may help clarify things https://en.wikipedia.org/wiki/Linear_equation – Giuseppe Oct 17 '16 at 13:24

score 7 · Accepted Answer · edited Oct 13 '16 at 20:00

7

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, $$x\approx7.267$$

edited Oct 13 '16 at 20:00

Em.

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answered Oct 13 '16 at 18:00

Ax.

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This should be it if I look at how it's worked out. – P.Yntema Oct 16 '16 at 18:43

mvw · Answer 3 · 2016-10-15T13:02:08.937

6

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 $f(B)$:

(Links to larger versions: left, right)

edited Oct 15 '16 at 13:02

answered Oct 13 '16 at 15:12

mvw

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I believe that is the incorrect approximate value of the solution. – Ian L Oct 13 '16 at 15:48
I see now that you used value e as your base. The solution, however, is still wrong. – Ian L Oct 13 '16 at 15:53
Ah yes. I separated it now. – mvw Oct 13 '16 at 17:23
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If B is not specified, it automatically means that it is 10. – P.Yntema Oct 16 '16 at 18:44
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. – Ian L Oct 20 '16 at 19:30

score 2 · Answer 4 · answered Oct 13 '16 at 15:08

2

Fill in details after you check the basic properties of logarithms, and assuming $\;\log=\log_{10}\;$:

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

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

answered Oct 13 '16 at 15:08

DonAntonio

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A term has disappeared – egreg Oct 13 '16 at 15:25
@egreg I don't think so....which one? – DonAntonio Oct 13 '16 at 15:58
Oh, assuming $\log=\log_{10}$. Why? – egreg Oct 13 '16 at 16:03
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@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. – DonAntonio Oct 13 '16 at 16:25
Yeah, whatever happened to the "standard" notation of ln for natural logs and lc for "common," or log-base 10 ? :-( – Carl Witthoft Oct 13 '16 at 19:54
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@CarlWitthoft My "standard" notation in high school was ln for natural logs, and log for base 10 logs. – Brian J Oct 13 '16 at 20:08
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@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. – DonAntonio Oct 13 '16 at 20:40
@DonAntonio It may be a low-level question, but I couldn't find the answer anywhere online. Therefor I think it was reasonable to post it here. – P.Yntema Oct 16 '16 at 18:47
@P.Yntema Please do note that elementary $;\neq;$ low-level. The problem is that in this site we cannot know what lever the asker has, so by "elementary" in this case I mean that this seems like a high school level question. – DonAntonio Oct 16 '16 at 19:36

Multiplying two logarithms

4 Answers4