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Attempting to solve the following equation, I got multiple different answers. Can someone explain what I did wrong and show me how to solve the problem correctly using the same method? Thanks

$$\huge\log (x^{\log x})=4$$


Note: $\log$ used in this question is $\log_{10}$

"Solution" 1:

Exponentiating, I get $$\large10^{\log x^{\log x}}=10^4$$ which simplifies to $$\large \log x=10000$$Solving for $x$, I get the answer of $$x=10^{10000}\quad\text{(obviously wrong)}$$

"Solution" 2:

Using the logarithmetic power rule, I simplify this equation to $$\large(\log x)^2=4$$ which simplifies to $$\large\log x=\pm 2$$ Solving these two equations, I get that $$x=100\;\text{or}\;x=\dfrac1{100}$$

Wolfram and Symbolab agree with the second solution, while Mathway gives a result of $x=10000$.

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    $\begingroup$ base 10 log $\!$ $\endgroup$ – suomynonA Feb 17 '17 at 4:34
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    $\begingroup$ Is it $(\log x)^{\log x}$ or $\log\left(x^{\log x}\right)$? $\endgroup$ – Thomas Andrews Feb 17 '17 at 4:49
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    $\begingroup$ @ThomasAndrews The second one $\endgroup$ – suomynonA Feb 17 '17 at 5:13
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    $\begingroup$ People just randomly hand out downvotes at no cost to perfectly valid questions. Maybe next time I'll just not include context; the question gets DOWNVOTED anyway ffs $\endgroup$ – suomynonA Feb 17 '17 at 5:34
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    $\begingroup$ @suomynonA It is good you ask such questions as it helps clarify the doubts of you and hopefully of the future readers. Logarithms are always a a little hard to understand at the first go ..... so your question is totally genuine (+unity ). You may want to check out my answer.... :) $\endgroup$ – user399078 Feb 17 '17 at 5:54
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EDIT : I assume that you mean $\log(x^{\log(x)})$.

Whenever you take the logarithm of something which is of the form $x^a$ then

$$\log(\lambda^a) \equiv a \log(\lambda).\,\,\,(♦)$$

What I mean to say is that the exponent "comes down".

Taking $\log(x^{\log(x)})$ and comparing it with $(♦)$, we get $\lambda = x$ and $a = \log(x)$. So the next steps should be clear :

$$\log(x^{\log(x)}) = (\log(x))^2 =4$$

$$\implies \log(x) = \pm 2$$

$$x = 10^2 \,\,\,\,\,\,\,OR \,\,\,\,\,x=10^{-2} = \dfrac{1}{100}$$

So your "second" solution is correct except for the first step where you wrote $LHS=2$ instead of $4$.

($LHS\,\,\equiv $ "Left Hand Side" (of the equation) )

Now coming to your first solution :

Note the following property :

$$b^{\log_b(\lambda)} \equiv \lambda\,\,\,\,.(♣)$$

That is, if both the "number to be raised to some power" and "base of logarithm in the exponent" are same, then what we get is "the number/quantity whose logarithm is being taken in the exponent".

Come to the original equation : $\log_{10}(x^{\log(x)}) =4$.

Now raise both sides of the equation to $10$ :

$$10^{\log_{10}(x^{\log(x)})} = 10^4$$

Compare this with $(♣)$ to get : $b=10$ and $\lambda = x^{\log(x)}$. So what we should get is :

$$x^{\log(x)} = 10000$$

That seems to bring us nowhere. Though it is possible to continue from here, the main point I want to highlight is the mistake you have committed in your "first" solution.

I have tried to make it as detailed as possible.

Hope this helps ! :-)

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    $\begingroup$ Yes, it made things clearer=) $\endgroup$ – suomynonA Feb 17 '17 at 6:18
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    $\begingroup$ @suomynonA Okay .... Good Luck with these !!! :-) $\endgroup$ – user399078 Feb 17 '17 at 6:18
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In the solution 1, how do you get the second equality? In fact, $$10^{\log x^{\log x}}=x^{\log x},$$ and the equality becomes $$x^{\log x}=10^4 \Leftrightarrow (\log x)^2=4.$$

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    $\begingroup$ How did you get $(\log x)^2=4$ from $x^{\log x}=10^4$? $\endgroup$ – suomynonA Feb 17 '17 at 4:41
  • $\begingroup$ just taking base 10 logarithm both sides :) $\endgroup$ – ntt Feb 17 '17 at 4:43
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    $\begingroup$ So you un-exponentiated and switched to what I did in solution 2 lol $\endgroup$ – suomynonA Feb 17 '17 at 4:44
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    $\begingroup$ Actually, $10^{\log x^{\log x}}\neq x^{\log x}$. $(10^{\log x})^{\log x}=x^{\log x}$, but that is not what you want. You can't use this technique to get a solution. $\endgroup$ – Thomas Andrews Feb 17 '17 at 4:47
  • $\begingroup$ I guess it depends on whether OP means $(\log x)^{\log x}$ or $\log\left(x^{\log x}\right)$. $\endgroup$ – Thomas Andrews Feb 17 '17 at 4:51

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