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What number do I need to multiply both sides with? I have worked for an hour on this but it is the first time I am using this website so it is impossible for me to write what I have already done. If you can give me tips on how to solve it I would appreciate it a lot!

$$ 2 \log_6 (\sqrt{x} + \sqrt[4]{x}) = \log_4 x $$

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  • $\begingroup$ Thanks for adding the parentheses. Forgot about those! $\endgroup$ – Brian Tung Jan 8 '16 at 20:03
  • $\begingroup$ Maybe use this: $\log_a x = (\log_b x) /( \log_b a)$. Convert everything to the same base. $\endgroup$ – GEdgar Jan 8 '16 at 20:05
  • $\begingroup$ how could this be useful? $\endgroup$ – Dr. Sonnhard Graubner Jan 8 '16 at 20:12
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Note that $2\log_b u = \log_b u^2$ for any $b$ and $u$, so that we can rewrite this as

$$ \log_6 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right) = \log_4 x $$

From here, the easiest thing is to guess, and observe that for the argument of the left-side log comes out nicely if $x$ is a fourth power. For instance, we might guess $x = 2^4 = 16$, which yields $x+2\sqrt[4]{x^3}+\sqrt{x} = 16+2(8)+4 = 36$, and indeed

$$ \log_6 36 = 2 = \log_4 16 $$

ETA: That this is the only solution for $x > 0$ can be seen if we write (following GEdgar's comments to the original post)

$$ \frac{\log_4 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right)}{\log_4 6} = \log_4 x $$

$$ \frac{\log_4 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right)}{\log_4 x} = \log_4 6 $$

or equivalently

\begin{align} \log_4 6 & = \log_x \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right) \\ & = 1 + \log_x \left(1+\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt{x}}\right) \\ & = 1 + 2\log_x \left(1+\frac{1}{\sqrt[4]{x}}\right) \end{align}

Note that the argument of the $\log_x$ is a decreasing function of $x$, which means that the entire right-hand side is a decreasing function of $x$ for $x > 1$, which in turn means that the solution $x = 16$ is the only solution for $x > 1$.

For $0 < x < 1$, the argument of the $\log_x$ is greater than $1$, but its base $x$ is less than $1$, so the log is negative, and the right-hand side is less than $1$, whereas the left-hand side is greater than $1$, so there are no solutions in that range, either.

Finally, $x = 1$ is not a solution of the original equation, by inspection.

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  • $\begingroup$ Thanks!!! I had rewritten it exactly like that but I hadn't even thought about guessing the answer logically!! I was expecting some other steps and a clear answer at the end. Brilliant! $\endgroup$ – user303668 Jan 8 '16 at 20:14
  • $\begingroup$ There might be some other way to this solution, and there might be a reasonably simple argument to show that $x = 16$ is the only solution. Still thinking about that. $\endgroup$ – Brian Tung Jan 8 '16 at 20:21
  • $\begingroup$ @user303668: OK, added some arguments to show the solution is unique. $\endgroup$ – Brian Tung Jan 8 '16 at 20:39
  • $\begingroup$ Thank you again for your time and explicit explanation. Although I prefer the first way, the second one is the "official solution" I will be using in questions like that in the future. At least, trying to... Thanks again $\endgroup$ – user303668 Jan 8 '16 at 20:49
  • $\begingroup$ Well, the second part is not a solution, so much as it shows (somewhat informally) that the solution we obtained in the first part is unique. $\endgroup$ – Brian Tung Jan 8 '16 at 20:50

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