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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.

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  • $\begingroup$ BTW, the logarithmic rule that would make this easiest is $\log_a b = 1/(\log_b a)$. Not what you asked for, but it's good to know. $\endgroup$ – JonathanZ supports MonicaC Dec 11 '20 at 5:13
  • $\begingroup$ According to 3Blue1Brown's video called Triangle of Power, the $3$ and $4$ get O-plussed, which gives $\frac{1}{1/3+1/4} = \frac{12}{7}$. $\endgroup$ – Toby Mak Dec 11 '20 at 5:21
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    $\begingroup$ Don't try to figure out what $x$ is in terms of $ab$. Try to figure out what $ab$ is in terms of $x$. $x = a^3$ so $a = x^{\frac 13}$ and $x^4 = b$ so $b= x^{\frac 14}$ so $ab = x^{\frac 13 + \frac 14}$. .... and from there you get $(ab)^{\frac 1{\frac 13 + \frac 14}} = x$. $\endgroup$ – fleablood Dec 11 '20 at 5:38
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$x = a^3 \Rightarrow a = x^{1/3}$ and $x = b^4 \Rightarrow b = x^{1/4}$

So, $ab = x^{\frac13 +\frac14} = x^{\frac7{12}} \Rightarrow x = (ab)^{12/7} \Rightarrow \log_{ab}(x) = \frac{12}7$

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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

$$ \dfrac{\log x}{\log a+ \log b} = \dfrac{\log x}{\log x^\frac13+ \log x^\frac14} $$

$$ = \dfrac{\log x}{\frac13 \log x+ \frac14\log x} = \dfrac{1}{\dfrac13+\dfrac14} = \dfrac{12}{7}$$

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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 $\log_{ab} x = \frac 1{\frac 13 + \frac 14}$.

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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 $\log_{ab}x=\dfrac{12}{7}$.

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$$a^3=x=b^4\implies b=a^{3/4}$$ Then $$\log_{ab}b=\log_{a^{7/4}}a^{3/4}=\frac{3/4}{7/4}=\frac37$$ and hence $\log_{ab}x=\frac{12}7$.

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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 $\log x$

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$$\log_{ab} x = \frac{1}{\log_x ab} = \frac{1}{\log_x a + \log_x b} = \frac{1}{\frac{1}{\log_a x} + \frac{1}{\log_b x}} = \frac{1}{\frac{1}{3} + \frac{1}{4}} = \frac{12}{7}$$

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