Interesting problem in logarithms

I know this place isn't for math problems/homework, and believe me I've been trying for a long time to solve this problem (45 mins to 1 hour) and besides I think many would find this useful or at least interesting, so here it is :

If $\log_6 15 = a$ and $\log_{12} 18 = b$ what is $\log_{25}24$? (The answer must be in terms of a and b)

• I think this has been asked before (deja vu). I can't seem to find it after searching though... – AvZ Feb 14 '15 at 16:42
• Are you sure it isn't $\log 24^{25}$ ? That would make a lot more sense. – DanielV Feb 14 '15 at 16:43
• @DanielV 100% positive – Ashkan Feb 14 '15 at 16:44
• @Ashkan Please edit your question to reflect your answer to surb's comment. If you need help with the latex, right click on his equation and click "Show Math As" -> "Tex Commands". – DanielV Feb 14 '15 at 16:45
• @DanielV The question seems fine to me, what part requires changing? – Ashkan Feb 14 '15 at 16:49

Use the identity

$$\log_x y=\frac{\ln y}{\ln x}$$

$$\frac{\ln 3+\ln 5}{\ln 2+\ln 3}=a$$

$$\frac{\ln 2+2\ln 3}{2\ln 2+\ln 3}=b$$

and you need to find

$$\frac{3\ln2+\ln 3}{2\ln 5}$$

in terms of $a$ and $b$.

We can simplify things a bit if we divide the numerators and denominators of all the fractions by $\ln 5$ and let $x=\frac{\ln 2}{\ln 5},\ y=\frac{\ln 3}{\ln 5}$. We then know that

$$\frac{y+1}{x+y}=a$$ $$\frac{x+2y}{2x+y}=b$$

and we want to find

$$\frac{3x+y}{2}$$

We now have two simultaneous equations with two variables. Simplifying the equations we get the linear equations

$$\begin{array}{rrrrr} ax & + & (a-1)y & = & 1 \\ (2b-1)x & + & (b-2)y & = & 0 \\ \end{array}$$

Using Cramer's rule, we get

$$x=\frac{2-b}{ab+a-2b+1}$$ $$y=\frac{2b-1}{ab+a-2b+1}$$

Substituting these into our formula, we get

$$\log_{25}24=\frac{5-b}{2ab+2a-4b+2}$$

This can be checked with a calculator, defining $a$ and $b$ as in your problem and calculating both my final expression and $\log_{25}24$. Those figures agree and are both $0.987317934353$.

• Thank you mate, I can proceed but I think it's better if you complete your answer – Ashkan Feb 14 '15 at 16:46
• To solve a fraction with 3 variables, two equations are enough. – Arashium Feb 14 '15 at 16:48
• Actually Rory I can't proceed and I'm frustrated as hell, could you please complete your answer, the curiosity is killing me – Ashkan Feb 14 '15 at 17:01
• This seems like an odd question. Surely you could write the result in terms of a and b with just: $$\frac{3\ln2+\ln 3}{2\ln 5}-a-b+a+b$$ How much trickery do you have to do before you can say that it's in terms of a and b? – Cruncher Feb 14 '15 at 17:36
• @Cruncher: A reasonable interpretation would be that the problem asks for the answer as a rational expression in $a$ and $b$. – Henning Makholm Feb 14 '15 at 17:57