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)
 A: Use the identity
$$\log_x y=\frac{\ln y}{\ln x}$$
Then your problem says that
$$\frac{\ln 3+\ln 5}{\ln 2+\ln 3}=a$$
$$\frac{\ln 2+2\ln 3}{2\ln 2+\ln 3}=b$$
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 $0.987317934353$.
