Solve the proof in problem involving logarithms

[chwala]

Homework Statement

If ##log_a n=x## and ##log_c n=y##, where ##n≠1##, prove that, $$\frac {x-y}{x+y}=\frac {log_b c-log_b a}{log_b c+log_ba}$$

Relevant Equations

understanding of logs

In my approach, i made use of change of base; i.e

$$x-y=\frac {log_b n}{log_b a} -\frac {log_b n}{log_b c}$$
$$x-y=\frac {log_b c ⋅log_b n - log_b n ⋅logba}{log_b a ⋅log_bc}$$
and
$$x+y=\frac {log_b n}{log_b a} +\frac {log_b n}{log_b c}$$
$$x-y=\frac {log_b c ⋅log_b n + log_b n ⋅logba}{log_b a ⋅log_bc}$$

$$→\frac {x-y}{x+y}=\frac {log_b n}{log_b n}⋅ \frac {log_b c - logba}{log_b c +log_b a}$$
$$=\frac {log_b c - logba}{log_b c +log_b a}$$

Any other way of looking at it...

 

[anuttarasammyak]

[tex]\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}[/tex]
where b is any positive number. So we get the formula of
[tex]\frac{1-y/x}{1+y/x}[/tex]
as requested.

 

[chwala]

[tex]\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}[/tex]
where b is any positive number. So we get the formula of
[tex]\frac{1-y/x}{1+y/x}[/tex]
as requested.

I do not seem to get what you did here...how did you arrive at this
[tex]\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}[/tex]

 

[anuttarasammyak]

Between the middle and the right terms, there is a term
[tex]\frac{\log_ba/\log_bn}{\log_bc/\log_bn}[/tex]