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 $x=10000$.

  • 1
    $\begingroup$ log here is natural log or base 10 log? $\endgroup$
    – User8976
    Feb 17 '17 at 4:32
  • 1
    $\begingroup$ base 10 log $\!$ $\endgroup$
    – suomynonA
    Feb 17 '17 at 4:34
  • 3
    $\begingroup$ Is it $(\log x)^{\log x}$ or $\log\left(x^{\log x}\right)$? $\endgroup$ Feb 17 '17 at 4:49
  • 1
    $\begingroup$ @ThomasAndrews The second one $\endgroup$
    – suomynonA
    Feb 17 '17 at 5:13
  • 2
    $\begingroup$ 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

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 ! :-)

  • 1
    $\begingroup$ Yes, it made things clearer=) $\endgroup$
    – suomynonA
    Feb 17 '17 at 6:18
  • 1
    $\begingroup$ @suomynonA Okay .... Good Luck with these !!! :-) $\endgroup$
    – user399078
    Feb 17 '17 at 6:18

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 $$x^{\log x}=10^4 \Leftrightarrow (\log x)^2=4.$$

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ So you un-exponentiated and switched to what I did in solution 2 lol $\endgroup$
    – suomynonA
    Feb 17 '17 at 4:44
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 17 '17 at 4:47
  • $\begingroup$ I guess it depends on whether OP means $(\log x)^{\log x}$ or $\log\left(x^{\log x}\right)$. $\endgroup$ Feb 17 '17 at 4:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.