Proving a logarithmic equation

[Aion]

Find [imath]\log_{54}{168}[/imath] if [imath]\log_7{12}=a[/imath] and [imath]\log_{12}{24}=b[/imath]

Solution attempt:

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

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

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

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

[math]\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[/math]
What do you guys think of my method? I'm not entirely sure why the cancellation of [imath]y[/imath] occurs, but I assume it because both the numerator and denominator are linear combinations of [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath].

 

[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 [imath] \log_2 [/imath], which saves the variable [imath] x [/imath] and I made no mistakes, then the linear equation system is



[math] \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}[/math]
and therefore



[math]\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}[/math]
This gives the same result.

 

[blamocur]

Find [imath]\log_{54}{168}[/imath] if [imath]\log_7{12}=a[/imath] and [imath]\log_{12}{24}=b[/imath]

Solution attempt:

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

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

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

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

[math]\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[/math]
What do you guys think of my method? I'm not entirely sure why the cancellation of [imath]y[/imath] occurs, but I assume it because both the numerator and denominator are linear combinations of [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath].

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