# Solve this logarithmic equation.

$2\log_6(x^{1/2}+x^{1/4}) = \log_4 x$

I don't have any good ideas how to start. Im stuck at this easy question, I don't know what is going on.

• Do you know the inverse function to $\log_n()$? It is exponentiating with $n$. It may also help you if you write the logs of every base back to $\ln()$ functions. ($\log_n(x) = \ln(x) / \ln(n)$) – Maximilian Gerhardt May 10 '16 at 19:51
• Wolfram does not find an exact solution. – zz20s May 10 '16 at 19:56
• Is there an exact solution? – user242559 May 10 '16 at 20:14
• @user242559 It's easy to verify that $x=16$ is a solution. The equation can be solved by hand with no trouble, but it seems a bit hefty for an "easy" precalculus exercise (unless there's a typo). – Corellian May 10 '16 at 20:18
• I feel so confused, I hate this kind of exercises. – semeduk May 10 '16 at 20:25

If $x = t^4$ (and $t > 0$) this becomes $$\dfrac{2 (\log(t)+\log(t+1))}{\log 6}=\dfrac{2 \log(t^2+t)}{\log(6)} = \dfrac{4 \log(t)}{\log(4)}$$ and you can isolate $\log(t+1)$: $$\log(t+1) = \left(\dfrac{\log(6)}{\log{2}}-1\right) \log(t) = \dfrac{\log(3)}{\log(2)} \log(t)$$ i.e. $$\dfrac{\log(t+1)}{\log(t)} = \dfrac{\log(2+1)}{\log(2)}$$ From this it's clear that $t=2$ (i.e. $x=16$) is a solution.
If $t > 2$, we have $t + 1 < (3/2) t$, so $\log(t+1) < \log(3/2) + \log(t)$, and then $$\dfrac{\log(t+1)}{\log(t)} < \dfrac{\log(3/2)}{\log(t)} + 1 < \dfrac{\log(3/2)}{\log(2)} + 1 = \dfrac{\log(3)}{\log(2)}$$ If $1 < t < 2$, then $t + 1 > (3/2) t$, and $\log(t+1) > \log(3/2) + \log(t)$, and then $$\dfrac{\log(t+1)}{\log(t)} > \dfrac{\log(3/2)}{\log(t)} + 1 > \dfrac{\log(3/2)}{\log(2)} + 1 = \dfrac{\log(3)}{\log(2)}$$ And of course, the equation doesn't work for $t \le 0$ or $t=1$, while if $0 < t < 1$ then $\log(t+1)/\log(t) < 0$. So $x=16$ is the only solution.
We move on to base $2$. $$\log_6(\sqrt x+\sqrt x)^2=\frac{\log_2(\sqrt x+\sqrt x)^2}{\log_2 6}$$ $$\log_4 x=\frac{\log_2 x}{\log_2 4}$$ Hence $$\frac{\log_2(\sqrt x+\sqrt x)^2}{\log_2 6}= \frac{\log_2 x}{\log_2 4}\iff \frac{\log_2(\sqrt x+\sqrt x)}{ \log_2 x}=\frac{\log_2 6}{4}$$ It is apparent that $x = 16$ satisfies this last equation so then we avoid all other kinds of calculation. The answer is $\color{red}{x=16}$