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The question:

$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \log_3 x \cdot \log_4 x \cdot \log_5 x \cdot \log_5 x \cdot \log_4 x \cdot \log_3 x$$

My mother who's a math teacher was asked this by one of her students, and she can't quite figure it out. Anyone got any ideas?

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    $\begingroup$ It would seem that three of the factors on each side are redundant and could simply be "cancelled". So is the question actually "Find $x$ such that $\log_3 x \cdot \log_4 x \cdot \log_5 x = 1 $" ? $\endgroup$ May 2, 2013 at 18:02
  • $\begingroup$ As noted by anorton in his answer, the factors can only be "cancelled" if $\log_m x \neq 0$ for $m = 3, 4, 5$, i.e., if $x \neq 1$. But $x = 1$ is also a solution. $\endgroup$
    – TMM
    May 2, 2013 at 18:16
  • $\begingroup$ @RecklessReckoner: this contradicts your comment to anorton. You can only cancel them if they are not zero. $\endgroup$ May 2, 2013 at 18:17
  • $\begingroup$ I admit I had neglected the possibility that all the factors might be zero. It struck me that there was an answer other than 1... $\endgroup$ May 2, 2013 at 18:18

5 Answers 5

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Use the identity $$ \log_a x=\ln x/\ln a. $$

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Following up on Jaeyong Chung's answer, and working it out:

$$ 1 =\log_3x\log_4x\log_5x$$ $$1=\frac{(\ln x)^3}{\ln3\ln4\ln5}$$ $$(\ln x)^3 = \ln3\ln4\ln5$$ $$(\ln x) = \sqrt[3]{\ln3\ln4\ln5}$$ $$x = \exp\left(\sqrt[3]{\ln3\ln4\ln5}\right) \approx 3.85093$$

EDIT: And, of course, the obvious answer that everyone will overlook: $x=1$ makes both sides of the equation zero. :D

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    $\begingroup$ ...which is about $3.85093$ according to Mathematica. $\endgroup$
    – TMM
    May 2, 2013 at 18:11
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    $\begingroup$ $x = 1$ is probably the answer the student in class was expecting... (We go in for non-trivial solutions around here.) $\endgroup$ May 2, 2013 at 18:15
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Given:

$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \log_3 x \cdot \log_4 x \cdot \log_5 x \cdot \log_5 x \cdot \log_4 x \cdot \log_3 x$$

One immediately obvious solution is $x = 1$. Regardless of the base $b$, $\log_b 1 = 0$. So $x = 1$ is a solution: it nullifies all factors simultaneously, making the equation true.

So on to chasing other solutions.

Firstly, note the repeating factors in right side, which condense to a square term:

$$\log_3 x \cdot \log_4 x \cdot \log_5 x = \left(\log_3 x \cdot \log_4 x \cdot \log_5\right)^2$$

Substitute z for the repeated subexpression: let $z = \log_3 x \cdot \log_4 x \cdot \log_5 x$. We then get a simplified form which clarifies the relationship:

$$z = z^2$$

This quadratic has two solutions:

$$z \in \lbrace 0, 1 \rbrace$$

Which corresponds to these two cases when we substitute back the original log factors for $z$:

$$\log_3 x \cdot \log_4 x \cdot \log_5 x \in \lbrace 0, 1 \rbrace$$

But the zero case corresponds to the $x = 1$ solution we already know, so henceforth we only care about the second case:

$$\log_3 x \cdot \log_4 x \cdot \log_5 x = 1$$

We can convert the logs to a common base, arbitrarily picking 3:

$$\log_3 x \cdot \frac{\log_3 x}{\log_3 4} \cdot \frac{\log_3 x}{\log_3 5} = 1$$

$$\frac{\left(\log_3x\right)^3}{\log_3 4\cdot \log_3 5} = 1$$

$$\left(\log_3x\right)^3 = {\log_3 4\cdot \log_3 5}$$

$$\log_3x = \left(\log_3 4\cdot \log_3 5\right)^{1/3}$$

$$x = 3^{\left(\log_3 4\cdot \log_3 5\right)^{1/3}}$$

This is approximately $3.8509$.


Appendix:

If the aim is to get a decimal figure with a calculator, it's better to use base 10 as the common base rather than 3, and this base is also better than $e$. We can then use a calculator which provides only a base 10 log function, and an $x^y$ button, which are more common than support for natural log, and a base $e$ exp function, or the availability of $e$ as a constant. Below, it is to be understood that $\log$ refers to $\log_{10}$:

$$\frac{\log x}{\log 3}\cdot\frac{\log x}{\log 4} \cdot \frac{\log x}{\log 5} = 1$$

$$\left(\log x\right)^3 = \log 3\cdot\log 4\cdot \log 5$$

$$\log x = \left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}$$

$$x = 10^{\left(\log 3\cdot\log 4\cdot \log 5\right)^{1/3}}$$

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As above, use $$log_a x=\frac{\ln x}{\ln a}$$ Hence, $$\log_3 x \cdot \log_4 x \cdot \log_5 x = 1$$ would become $$\frac{\ln x \cdot \ln x \cdot \ln x}{\ln 3 \cdot \ln 4 \cdot \ln5}=1$$

Thus, $$(\ln x)^3=\ln 3 \cdot \ln 4 \cdot \ln 5=2.45117$$

Hence, $$\ln x = 1.34831$$

This gives $x$ as $$x=e^{1.34831}=3.85091$$ approximately.

EDIT: Here, we assume that $x \neq 1$. So $x=1$ is also a possible solution.

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    $\begingroup$ $=$ is not the same as $\approx$ $\endgroup$
    – Ruslan
    May 2, 2013 at 19:25
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The "change-of-base" formula should at least allow you to reduce this to

$$\log_3 x \cdot \left(\frac{\log_4 x}{\log_4 3}\right) \cdot \left(\frac{\log_5 x}{\log_5 3}\right) \ = \ (\log_3 x)^3 \ = \ \frac{1}{\log_4 3 \cdot \log_5 3} . $$

The number won't be "pretty"...

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