# 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.

• 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. – JonathanZ supports MonicaC Dec 11 '20 at 5:13
• According to 3Blue1Brown's video called Triangle of Power, the $3$ and $4$ get O-plussed, which gives $\frac{1}{1/3+1/4} = \frac{12}{7}$. – Toby Mak Dec 11 '20 at 5:21
• 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 $(ab)^{\frac 1{\frac 13 + \frac 14}} = x$. – fleablood Dec 11 '20 at 5:38

$$x = a^3 \Rightarrow a = x^{1/3}$$ and $$x = b^4 \Rightarrow b = x^{1/4}$$

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

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

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

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

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

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

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

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 $$\log x$$

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