<s>[INST] 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?
 [/INST] 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
$$\log_{24}(48)= \frac{\log_{12}(48)}{\log_{12}(24)}=\frac{2\log_{12}(2)+1}{\log_{12}(2)+1}$$
$$=\frac{2-k+1}{2-\frac{k}{2}}=\frac{6-2k}{4-k}.$$
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