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If $$\log_{10}(x)\log_{10}(2) = 2$$ What is $x$ ?

WolframAlpha says $x = e^{\frac2{\log_{10}(2)}}$

But i don't understand why it is.. Please explain it. Thanks

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  • $\begingroup$ It would help to know exactly what you wrote as your input to WA, because the answer you say WA gave is not the answer to this question. $\endgroup$
    – Thomas Andrews
    Apr 22, 2015 at 16:20

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There's really not much to it, but you probably used $\ln$ when asking W|A (log is interpreted as the natural logaritm by default). $$\log_{10} x \log_{10} 2 = 2 \\ \Rightarrow \log_{10} x = \frac2{\log_{10}2} \\ \Rightarrow x = 10^{\log_{10} x} = 10^{\frac2{\log_{10}2}}$$

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That's because, when you write down $\log$ in WolframAlpha, it is interpreted as $\log_e$ or $\ln$. So WolframAlpha solved the equation

$$\ln(x)\ln(2)=2$$

Which is solved by first dividing the equation by $\ln 2$, obtaining $$ln(x) = \frac{2}{\ln 2}$$

Then using the fact that $a=b\iff e^a=e^b$ to get

$$(x=)e^{\ln x} = e^{\frac{2}{\ln2}}$$

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  • $\begingroup$ Except, oddly, WA seems to get the $\log_{10}(2)$ right - it doesn't interpret it as $\ln 2$. $\endgroup$
    – Thomas Andrews
    Apr 22, 2015 at 16:15
  • $\begingroup$ @ThomasAndrews No, WA writes log, which is the natural log. See wolframalpha.com/input/?i=solve+log%28x%29log%282%29%3D2 for proof. $\endgroup$
    – 5xum
    Apr 22, 2015 at 16:16
  • $\begingroup$ This is what the OP wrote was WA's answer: $$x = e^{\frac2{\log_{10}(2)}}$$ That's the correct answer to $\log(x)\log_{10}(2)=2$. $\endgroup$
    – Thomas Andrews
    Apr 22, 2015 at 16:18
  • $\begingroup$ @ThomasAndrews True, but since (if you only followed the link I provided) WA returns $\frac{2}{\log (2)}$ without explicitly saying that it is $\log_2$, and since it is obvious that the OP did not know that WA interprets $\log$ as $\log_e$ and not $\log_{10}$, it is safe to assume that the OP, when reading WA's result, interpreted $\log$ as $\log_{10}$. $\endgroup$
    – 5xum
    Apr 22, 2015 at 16:28
  • $\begingroup$ So, in other words, you are not explaining to the OP why he got the answer he got from WA, nor do you mention that your answer is different, nor do you answer the original question. $\endgroup$
    – Thomas Andrews
    Apr 22, 2015 at 16:29
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It's just a rearrangement of variables.

So, say:

  • a = log(x)
  • b = log(2)
  • c = 2

Then we have a*b = c, which means a=c/b

To get x by itself simply take the exponential of both sides (because the exponential of the log of x is just x).

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