Proving a logarithmic equation

[Aion]

Find $\log_{54}{168}$ if $\log_7{12}=a$ and $\log_{12}{24}=b$

Solution attempt:

We first rewrite the given logarithms with the change of base formula

$$\log_{54}{168}=\frac{\ln{168}}{\ln54}=\frac{3\ln2+\ln3+\ln7}{\ln2+3\ln3}$$
and
$$a=\log_{7}{12}=\frac{\ln12}{\ln7}=\frac{2\ln2+\ln3}{\ln7}$$$$b=\log_{12}{24}=\frac{\ln24}{\ln12}=\frac{3\ln2+\ln3}{2\ln2+\ln3}$$
Now define $$x=\ln2,y=\ln3,z=\ln7$$
From the two conditions above on $a$ and $b$, we get the two linear equations
$$\begin{aligned} 2x+y-az=0 \\ (2b-3)x+(b-1)y=0 \end{aligned}$$
Divide each term by $z$ such that $x'=\frac{x}{z}$, $y'=\frac{y}{z}$. This gives us the system
$$\begin{aligned} 2x'+y'-a=0 \\ (2b-3)x'+(b-1)y'=0 \end{aligned}$$
Using the cross-multiplication rule, we find that

$$\frac{x'}{a(b-1)}=\frac{y'}{a(3-2b)}=1$$
which shows us that
$$\frac{x'}{y'}=\frac{x}{y}=\frac{b-1}{3-2b}\implies x=\frac{(b-1)y}{3-2b}$$
We now solve for $z$ in terms of $y$.

$$z=\frac{2x+y}{a}=\frac{2\left(\frac{(b-1)y}{3-2b}\right)+y}{a}=\frac{y}{a(3-2b)}$$
Now we can substitute the values for $x$ and $z$ back into the expression for $\log_{54}{168}$ such that the $y$ terms cancel.

$$\log_{54}{168}=\frac{3x+y+z}{3y+x}=\frac{3\left(\frac{(b-1)y}{3-2b}\right)+y+\left(\frac{y}{a(3-2b)}\right)}{3y+\left(\frac{(b-1)y}{3-2b}\right)}=\frac{b+\frac{1}{a}}{8-5b}=\frac{ab+1}{a(8-5b)}\quad\square$$
What do you guys think of my method? I'm not entirely sure why the cancellation of $y$ occurs, but I assume it because both the numerator and denominator are linear combinations of $x$, $y$, and $z$.

 

[fresh_42]

I haven't checked all your steps, but it looks fine. I only think there may be a shorter version using the fact that 12 is almost everywhere in the problem statement.

 

[fresh_42]

If we work with $\log_2$, which saves the variable $x$ and I made no mistakes, then the linear equation system is



$$\begin{pmatrix}1&-a\\b-1&0 \end{pmatrix}\begin{pmatrix}\log_2 3\\\log_2 7\end{pmatrix}=\begin{pmatrix}-2\\3-2b\end{pmatrix}\\[12pt] \begin{pmatrix}\log_2 3\\ \log_2 7\end{pmatrix}=\dfrac{1}{ab-a}\begin{pmatrix}0&a\\1-b&1\end{pmatrix}\begin{pmatrix}-2\\3-2b\end{pmatrix}$$
and therefore



$$\begin{array}{lll} \log_2 3&=\dfrac{3-2b}{b-1}\\[12pt] \log_2 7&=\dfrac{1}{ab-a}\\[12pt] \log_{54}168&=\dfrac{3+\log_2 3+\log_2 7}{1+\log_2 3}=\ldots \end{array}$$
This gives the same result.

 

[blamocur]

Find $\log_{54}{168}$ if $\log_7{12}=a$ and $\log_{12}{24}=b$

Solution attempt:

We first rewrite the given logarithms with the change of base formula

$$\log_{54}{168}=\frac{\ln{168}}{\ln54}=\frac{3\ln2+\ln3+\ln7}{\ln2+3\ln3}$$
and
$$a=\log_{7}{12}=\frac{\ln12}{\ln7}=\frac{2\ln2+\ln3}{\ln7}$$$$b=\log_{12}{24}=\frac{\ln24}{\ln12}=\frac{3\ln2+\ln3}{2\ln2+\ln3}$$
Now define $$x=\ln2,y=\ln3,z=\ln7$$
From the two conditions above on $a$ and $b$, we get the two linear equations
$$\begin{aligned} 2x+y-az=0 \\ (2b-3)x+(b-1)y=0 \end{aligned}$$
Divide each term by $z$ such that $x'=\frac{x}{z}$, $y'=\frac{y}{z}$. This gives us the system
$$\begin{aligned} 2x'+y'-a=0 \\ (2b-3)x'+(b-1)y'=0 \end{aligned}$$
Using the cross-multiplication rule, we find that

$$\frac{x'}{a(b-1)}=\frac{y'}{a(3-2b)}=1$$
which shows us that
$$\frac{x'}{y'}=\frac{x}{y}=\frac{b-1}{3-2b}\implies x=\frac{(b-1)y}{3-2b}$$
We now solve for $z$ in terms of $y$.

$$z=\frac{2x+y}{a}=\frac{2\left(\frac{(b-1)y}{3-2b}\right)+y}{a}=\frac{y}{a(3-2b)}$$
Now we can substitute the values for $x$ and $z$ back into the expression for $\log_{54}{168}$ such that the $y$ terms cancel.

$$\log_{54}{168}=\frac{3x+y+z}{3y+x}=\frac{3\left(\frac{(b-1)y}{3-2b}\right)+y+\left(\frac{y}{a(3-2b)}\right)}{3y+\left(\frac{(b-1)y}{3-2b}\right)}=\frac{b+\frac{1}{a}}{8-5b}=\frac{ab+1}{a(8-5b)}\quad\square$$
What do you guys think of my method? I'm not entirely sure why the cancellation of $y$ occurs, but I assume it because both the numerator and denominator are linear combinations of $x$, $y$, and $z$.

I've checked your answer with a quick and dirty script, and it looks correct.