Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?

You can convert all to the smallest common base $2$:
$$\log_{12}36=\frac{\log_{2}36}{\log_2 12}=\frac{2+2\log_2 3}{2+\log_2 3}=k  \Rightarrow \log_2 3=\frac{2k-2}{2-k}.$$
Hence:
$$\log _{24}48=\frac{\log_2 48}{\log_2 24}=\frac{4+\log_23}{3+\log_23}=\frac{4+\frac{2k-2}{2-k}}{3+\frac{2k-2}{2-k}}=\frac{6-2k}{4-k}.$$
