# Solving logarithmic equation, different bases

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!

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

• Thanks for adding the parentheses. Forgot about those! Jan 8, 2016 at 20:03
• Maybe use this: $\log_a x = (\log_b x) /( \log_b a)$. Convert everything to the same base. Jan 8, 2016 at 20:05
• how could this be useful? Jan 8, 2016 at 20:12

Note that $2\log_b u = \log_b u^2$ for any $b$ and $u$, so that we can rewrite this as

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

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 $x+2\sqrt{x^3}+\sqrt{x} = 16+2(8)+4 = 36$, 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)

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

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

or equivalently

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

• 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! Jan 8, 2016 at 20:14
• 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. Jan 8, 2016 at 20:21
• @user303668: OK, added some arguments to show the solution is unique. Jan 8, 2016 at 20:39
• 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 Jan 8, 2016 at 20:49
• 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. Jan 8, 2016 at 20:50