Solve the following:
In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$, for some fixed $b \geq 2$. A Martian student writes down
\begin{align*}3 \log(\sqrt{x}\log x) &= 56\\\log_{\log (x)}(x) &= 54
\end{align*}and finds that this system of equations has a single real number solution $x > 1$. Find $b$
My progress:
Let $\log_bx=c$.
So $\log_cx=54\implies x=c^{54}$. Also $$3\log(\sqrt{x}\log x) =56\implies 3\log(\sqrt{x})+3\log(x) =56\implies 3\log_b(\sqrt{x})+3c=56.$$
By base changing formula, we get $$\log_b(\sqrt{x})=\frac{\log_c{\sqrt{x}}}{\log_cb}=27\cdot \log_bc=27\cdot \frac{\log_xc}{\log_xb}=27\cdot \frac{1}{54}\cdot c=\frac{c}{2}.$$
So $$\frac{3c}{2}+3c=56\implies c=\frac{56\times 2}{9}.$$
But actually, $c=36$. So please help me find the mistake in my proof.
Best Answer
HINT
Using identity
$$\log_{~~q} ~p=\frac{\log p}{\log q}$$
for any base. Next recast and solve
$$ u=log \sqrt x, v=log ~log x~ $$
$$ u+v=\frac{56}{3},~ \frac{u}{v}=27. $$