Calculate $\log_{a}(ab)$, when $\log_{ab}(b)$ is known 
If you know that $\log_{ab}(b) = k$, calculate $\log_{a}(ab)$. 

Last time I was asked two times about this problem. $a,b$ was given, constant, such that $a,b \in \mathbb{Z} \wedge a,b > 1 \wedge \gcd(a,b) = 1$. What is strategy to solving this problem?
In fact it was a little bit harder - numbers occurred at different powers, eg. $\log_{25}50$, when $\log_{10}64 = k$, but it doesn't change a lot and this topic is about strategy to solve similar problem.
As someone asked again (example) I was a little bit confused, because problem is easy, and  I thought, I should make some .pdf, but maybe it helps someone here. Alternative solutions are welcome.
 A: If you know that $\log_{ab}(b) = k$ 
then $b= (ab)^k=a^k b^k$ and $b^{1-k}=a^k $, so $b=a^{k/(1-k)}$ and  $\log_{a}(b)=\frac{k}{1-k}$ 
leading to $\log_{a}(ab) = \log_{a}(a) + \log_{a}(b) = 1+\frac{k}{1-k}=\dfrac{1}{1-k}$. 
A: $$\log_aab=\log_aa+\log_ab=1+\log_ab=1+\frac{\log_{ab}b}{\log_{ab}a}\;\;(**)$$
and
$$\log_{ab}a=\frac{\log_aa}{\log_aab}=\frac1{\log_aab}$$
Substitute now back in (**) and get
$$\log_aab=1+\frac{\log_{ab}b}{\frac1{\log_aab}}=1+\log_aab\log_{ab}b\implies\log_aab=\frac1{1-\log_{ab}b}$$
A: Strategy should be clear. Logarithms have to have same bases. We will use the basic facts.
$$\begin{split}
&\left(\forall x,y \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\log_{x}{y} = \frac{1}{\log_{y}{x}} \right)\\
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall y,z \in \mathbb{R}^{+} \right)
\left(\log_{x}{yz} = \log_{x}{y} + \log_{x}{z} \right)\\
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\log_{x}{x} = 1 \right)
\end{split}$$
It is easer to change known logarithm.
$$\begin{align*}\begin{split}
k = \log_{ab}b = \frac{1}{\log_{b}ab} = \frac{1}{\log_{b}a + \log_{b}b} =
\frac{1}{\log_{b}a + 1} &= k &\Longleftrightarrow\\
k(\log_{b}a + 1) &= 1  &\Longleftrightarrow \\
\log_{b}a &= \frac{1-k}{k} &\Longleftrightarrow\\
\frac{1}{\log_{a}b} &= \frac{1-k}{k} & \Longleftrightarrow\\
\log_{a}{b} &= \frac{k}{1-k}
\end{split}\end{align*}$$
Now we just have to transform searched logarithm.
$$ \log_a{ab} = \log_{a}{a} + \log_{a}{b} = 1 + \log_{a}{b} =
1 + \frac{k}{1-k} = \frac{1}{1-k}$$

And for given example. Now we have to use more facts.
$$\begin{split}
&\left(\forall x \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall \alpha \in \mathbb{R} \right)
\left(\log{x^{\alpha}} = \alpha \cdot \log{x} \right)\\
&\left(\forall x,y \in \mathbb{R}^{+}\smallsetminus\lbrace1\rbrace \right)
\left(\forall \alpha \in \mathbb{R} \right)
\left(\log_{x^{\alpha}}{y} = \frac{1}{\alpha} \cdot \log{y} \right)
\end{split}$$
$$\begin{align*}\begin{split}
k = \log64 = \log 2^{6} = 6 \log 2 = \frac{6}{\log_{2}(2 \cdot 5)} =
\frac{6}{\log_2{2} + \log_2{5}} =
\frac{6}{1 + \log_2{5}} &= k &\Longleftrightarrow\\
k(1 + \log_{2}5) &= 6 &\Longleftrightarrow\\
\log_{2}5 &= \frac{6-k}{k} &\Longleftrightarrow\\
\frac{1}{\log_{5}{2}} &= \frac{6-k}{k} &\Longleftrightarrow\\
\log_{5}{2} &= \frac{k}{6-k}
\end{split}\end{align*}$$
And transform searched logarithm $\log_{25}{50}$.
$$\begin{align*}\begin{split}
\log_{25}{50} = \log_{25}(25 \cdot 2) &= \log_{25}(25) + \log_{25}(2)\\
&= 1 + \log_{5^2}2 = 1 + \frac{1}{2}\log_{5}{2}\\
&= 1 + \frac{k}{2 \cdot (6-k)}\\
&= \frac{12-k}{12-2k} 
\end{split}\end{align*}$$
So the answer is $\log_{25}50 = \frac{12-k}{12-2k}$.
A: $$k=\log_{ab}b=\frac{\log b}{\log a+\log b}\iff \log b=\frac{1-k}k\cdot\log a$$
$$\implies\log_a{ab}=\frac{\log a+\log b}{\log a}=1+\frac k{1-k}=\frac1{1-k}$$
A: $\log_{ab}(b)=\log_{ab}\left(\dfrac{ab}{a}\right)=1-\log_{ab}(a)=1-\dfrac{1}{\log_{a}(ab)}$
A: Not really any new ideas or methods in here that aren't in other answers, but just thought this particular progression was quite simple and direct:
$1=\log_{ab}(ab)=\log_{ab}(a)+\log_{ab}(b)=\log_{ab}(a)+k=\frac{\log_{a}(a)}{\log_{a}(ab)}+k=\frac{1}{\log_{a}(ab)}+k$
Thus $\log_{a}(ab)=(1-k)^{-1}$.
A: Going back to logarithms in base $e$, $$\log_{ab}(b)=\frac{\log (b)}{\log (a b)}=k$$ $$\log_{a}(ab)=\frac{\log (a b)}{\log (a)}$$
I am sure that you can take from here.
A: Let $A=\log a; B=\log b$.
$$\begin{align}\log_{ab}b =\frac B{A+B}=k\qquad \cdots (1)\\
\log_a{ab}=\frac {A+B}A\qquad \cdots (2)\end{align}$$
From (1), apply $\frac D{D-N}$, where $D$=denominator, $N$=numerator:
$$\frac 1{1-k}=\frac {A+B}A$$
which equals (2), hence
$$\log_a{ab}=\frac 1{1-k}$$
