If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$? 
If $\log_ax=3$ and $\log_bx=4$, then what is $\log_{ab}x$?

I'm sure there is some logarithmic rule that can allow me to solve this in one or two steps, but as I am not very familiar with logarithms, I have decided to use the definition of logarithms to convert them to exponential equations.

So from $\log_ax=3$, we have $$x=a^3$$Similarly, from $\log_bx=4$, we have $$x=b^4$$Then I multiply those two equations together:$$\begin{align}x^2&=a^3b^4\\x^2&=(ab)^3\cdot b\\2\log_{ab}x&=3+\log_{ab}b\\\log_{ab}x&=\frac{3+\log_{ab}b}2\end{align}$$The only problem now is the extra $\log_{ab}b$ term. If I can get the value of that, then I will solve the problem. But right now I can't seem to find a way to complete this problem. Can anyone provide some insight on this problem? Thanks.
 A: $x = a^3 \Rightarrow a = x^{1/3}$  and $x = b^4  \Rightarrow b = x^{1/4}$
So, $ab = x^{\frac13 +\frac14} = x^{\frac7{12}} \Rightarrow x = (ab)^{12/7} \Rightarrow \log_{ab}(x) = \frac{12}7$
A: Just remember
$$ \log_vu= \dfrac{\log_p u}{\log_p v}$$
where $p$ can be any arbitrary real number. Writing as power to base
$$ x=a^3 = b^4$$
The given quantity
$$ \dfrac{\log x}{\log a+ \log b}  = \dfrac{\log x}{\log x^\frac13+ \log x^\frac14}  $$
$$ = \dfrac{\log x}{\frac13 \log x+ \frac14\log x} = \dfrac{1}{\dfrac13+\dfrac14} = \dfrac{12}{7}$$
A: Just go back to definitions.
$\log_a x = 3\iff a^3 = x$.
And $\log_b x = 4 \iff b^4=x$.
So $a = x^{\frac 13}$ and $b=x^{\frac 14}$.
So $ab = x^{\frac 13}x^{\frac 14}= x^{\frac 13 + \frac 14}$.
So $(ab)^{\frac 1{\frac 13 + \frac 14} } = x$.
Which means $\log_{ab} x = \frac 1{\frac 13 + \frac 14}$.
A: Actually from $x=a^3$ and $x=b^4$ we immediately have $a=b^{4/3}$. Thus $\log_{ab} b = \log_{b^{7/3}}b= \dfrac{3}{7}$. Thus we get $\log_{ab}x=\dfrac{12}{7}$.
A: $$a^3=x=b^4\implies b=a^{3/4}$$
Then
$$\log_{ab}b=\log_{a^{7/4}}a^{3/4}=\frac{3/4}{7/4}=\frac37$$
and hence $\log_{ab}x=\frac{12}7$.
A: As $\log_ab=\dfrac{\log b}{\log a}$ when all the logarithms remain defined,
$$3=\log_ax=\dfrac{\log x}{\log a}\implies\log a=?$$
Similarly, $\log b=\dfrac{\log x}4$
$$\log_{ab}x=\dfrac{\log x}{\log a+\log b}$$
Replace the values of $\log a,\log b$ in terms of $\log x$
A: $$\log_{ab} x = \frac{1}{\log_x ab} = \frac{1}{\log_x a + \log_x b} = \frac{1}{\frac{1}{\log_a x} + \frac{1}{\log_b x}} = \frac{1}{\frac{1}{3} + \frac{1}{4}} = \frac{12}{7}$$
