Calculate $\log_{a}(ab)$, when $\log_{ab}(b)$ is known

Question

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.

Answer 1

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 $\log_{a}(ab) = \log_{a}(a) + \log_{a}(b) = 1+\frac{k}{1-k}=\dfrac{1}{1-k}$.

Answer 2

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.

$$ \log_a{ab} = \log_{a}{a} + \log_{a}{b} = 1 + \log_{a}{b} = 1 + \frac{k}{1-k} = \frac{1}{1-k}$$

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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 $\log_{25}50 = \frac{12-k}{12-2k}$.

Answer 3

$$\log_aab=\log_aa+\log_ab=1+\log_ab=1+\frac{\log_{ab}b}{\log_{ab}a}\;\;(**)$$

and

$$\log_{ab}a=\frac{\log_aa}{\log_aab}=\frac1{\log_aab}$$

Substitute now back in (**) and get

$$\log_aab=1+\frac{\log_{ab}b}{\frac1{\log_aab}}=1+\log_aab\log_{ab}b\implies\log_aab=\frac1{1-\log_{ab}b}$$

Answer 4

$$k=\log_{ab}b=\frac{\log b}{\log a+\log b}\iff \log b=\frac{1-k}k\cdot\log a$$

$$\implies\log_a{ab}=\frac{\log a+\log b}{\log a}=1+\frac k{1-k}=\frac1{1-k}$$

Answer 5

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.

Answer 6

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

$$\log_a{ab}=\frac 1{1-k}$$

Answer 7

$\log_{ab}(b)=\log_{ab}\left(\dfrac{ab}{a}\right)=1-\log_{ab}(a)=1-\dfrac{1}{\log_{a}(ab)}$