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?

With $z = \log_2(3)$:
$$L = \log_{24}(48) = \frac{\ln(48)}{\ln(24)} = \frac{\ln(2^4\times 3)}{\ln(2^3 \times 3)} = \frac{4\ln(2) + \ln(3)}{3\ln(2) + \ln(3)} = \frac{4+z}{3 + z}$$
$$K = \log_{12}{(36)} = \frac{\ln(36)}{\ln(12)} = \frac{\ln(2^2\times 3^2)}{\ln(2^2 \times 3)} = \frac{2\ln(2) + 2\ln(3)}{2\ln(2) + \ln(3)} = \frac{2+2z}{2 + z}$$
Just solve for $z$ and plug:
$$z = \frac{2K-2}{K-2}$$
$$L = \frac{2K - 6}{K-4}$$
