Proof the expession $\log_{12}{18}\times\log_{24}{54} + 5(\log_{12}{18}-\log_{24}{54})=1$

Question

I am trying to proof the following expression (without a calculator of course).

$\log_{12}{18}\times\log_{24}{54} + 5(\log_{12}{18}-\log_{24}{54})=1$

I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of $\log_{12}{3}$ and $\log_{12}{2}$ but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong.

Thanks ;) ( if there are more levels to this task, I'd like a hint, not a complete solution)

Answer 1

Let $\log_{12}18=a, \, \log_{24}54=b$. So you want to prove that $\color{red}{ab+5(a-b)=1}$.

Then you get $12^a=18$ and $24^b=54$. Now factor everything in powers of $2$ and $3$ to get $$2^{2a-1}=3^{2-a} \qquad 2^{3b-1}=3^{3-b}.$$ From this taking $\log$ base $2$ you will get: $$2a-1 = (2-a)\log_23 \qquad 3b-1 = (3-b)\log_23.$$ Furthermore you get $$\frac{2a-1}{2-a}=\frac{3b-1}{3-b}.$$ Now simplify this and see you will get the expression written on the first line.

Answer 2

The follow-your-nose approach would be, in my opinion, \begin{align*} \log_{12}18 \cdot\log_{24}54 &+ 5(\log_{12}18-\log_{24}54) \\ &= \frac{\log18}{\log12} \frac{\log54}{\log24} +5 \bigg( \frac{\log18}{\log12} - \frac{\log54}{\log24} \bigg) \\ &= \frac{\log18}{\log12} \frac{\log54}{\log24} +5 \bigg( \frac{\log18 \cdot \log24 - \log54 \cdot \log12}{\log12 \cdot \log24} \bigg) \\ &= \frac{\log18\cdot\log54 + 5 \log18 \cdot \log24 - 5\log54 \cdot \log12}{\log12 \cdot \log24}. \end{align*} To save space, write $t=\log2$ and $h=\log3$. Then $\log18=\log2+2\log3=t+2h$, etc., and we get \begin{align*} &\frac{\log18\cdot\log54 + 5 \log18 \cdot \log24 - 5\log54 \cdot \log12}{\log12 \cdot \log24} \\ &\qquad{}= \frac{(t+2h)(t+3h) + 5(t+2h)(3t+h) - 5(t+3h)(2t+h)}{(2t+h)(3t+h)} \\ &\qquad{}= \frac{t^2+5th+6h^2 + 15t^2+35th+10h^2 - (10t^2+35th+15h^2)}{6t^2+5th+h^2} =1. \end{align*}