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?
 A: It seems easier to use $\log_{12}$ rather than $\log$.
$\log_{12}(36)=\log_{12}(3)+1=k$, and 
$$2\log_{12}(2)+\log_{12}(3)=\log_{12}(12)=1,$$ so $2\log_{12}(2)=2-k$,
or $$\log_{12}(2)=1-\frac{k}{2}$$
Thus
$$\log_{24}(48)= \frac{\log_{12}(48)}{\log_{12}(24)}=\frac{2\log_{12}(2)+1}{\log_{12}(2)+1}$$
$$=\frac{2-k+1}{2-\frac{k}{2}}=\frac{6-2k}{4-k}.$$
A: 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:
$$z = \frac{2K-2}{K-2}$$
$$L = \frac{2K - 6}{K-4}$$
A: We have
$$\log _{24}48=\frac{\log _{12}48}{\log _{12}24}=\frac{\log _{12}36+\log _{12}\frac43}{\log _{12}36+\log _{12}\frac23}=\frac{k+\log _{12}\frac43}{k+\log _{12}\frac23}$$
A: 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:
$$\log _{24}48=\frac{\log_2 48}{\log_2 24}=\frac{4+\log_23}{3+\log_23}=\frac{4+\frac{2k-2}{2-k}}{3+\frac{2k-2}{2-k}}=\frac{6-2k}{4-k}.$$
