Logarithms and More Logarithms!

From $\log_b x = 4$ , $\implies \dfrac {\log_a x}{\log_a b} = \dfrac 3{\log_a b} = 4$ , $\implies \color{#3D99F6} \log_a b = \dfrac 34$ . Then we have $\log_{ab} x = \dfrac {\log_a x}{\log_a (ab)} = \dfrac 3{\log_a a + \color{#3D99F6}\log_a b} = \dfrac 3{1+\color{#3D99F6} \frac 34} = \dfrac {12}7 \approx \boxed{1.714}$

We are given that $\log_a x = 3$ and that $\log_b x = 4$ . Our objective is to find what $\log_{ab}$ equals to. Well, to do this, we must first rewrite the given logarithms using a different form. This gives us that $\log_a x = 3 \implies \frac{\log(x)}{\log(a)}=3$ and that $\log_b x = 4 \implies \frac{\log(x)}{\log(b)}=4$ . We can see that both logarithmic function have $\log(x)$ as their numerators, which will be used later. Now, we can move on to finding what $\log_{ab} \ x$ is. Since $\log_{ab}$ is the same thing as $\log_a + log_b$ , we can create the following expression: $\frac{\log(x)}{\log_{ab}} \Rightarrow \frac{\log(x)}{\log_a + \log_b}$ . Now is the part where it gets somewhat tricky. We can use another rule for logarithms to get the following derived expressions for $\log_a x=3$ and $\log_b x=4$ - $\frac{\log(x)}{\frac{\log(x)}{3}+\frac{\log(x)}{4}} \implies \frac{1}{\frac{1}{3}+\frac{1}{4}} \Rightarrow \frac{1}{\frac{7}{12}} \Rightarrow \frac{12}{7} \approx \boxed{1.714}$

By the properties of logarithms, $\log_x a = \dfrac{1}{3}$ and $\log_x b = \dfrac{1}{4}.$ Thus,

$\begin{aligned} \log_{ab} x &= \dfrac{1}{\log_x ab} \\ &= \dfrac{1}{\log_x a + \log_x b} \\ &= \dfrac{1}{\frac{1}{3} + \frac{1}{4}} \\ &= \dfrac{12}{7} \\ &\approx \boxed{1.714}. \end{aligned}$

$\log_a{x} = 3 \implies a^3 = x$ , $\log_b{x} = 3 \implies b^3 = x$ , and $\log_{ab}{x} = y \implies (ab)^y = x$ .

If $x = a^3 = b^4$ , then $a = b^\frac{4}{3}$ .

If $a = b^\frac{4}{3}$ and $x = b^4$ , then $(ab)^y = x \implies (b^\frac{4}{3}b)^y = b^4 \implies b^\frac{7y}{3} = b^4 \implies \frac{7y}{3} = 4 \implies y = \frac{12}{7} \approx \boxed{1.714}$ .