Solve the proof in problem involving logarithms

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    Logarithms Proof
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The discussion revolves around solving a logarithmic proof using the change of base formula. The user presents a series of equations to express relationships between variables x, y, and logarithmic terms. They derive a formula for the ratio of y to x, leading to the expression (1 - y/x) / (1 + y/x). Another participant questions the derivation of the ratio y/x and points out a missing term in the transition between steps. The conversation highlights the complexity of manipulating logarithmic identities and the need for clarity in each transformation.
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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...
 
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\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}
where b is any positive number. So we get the formula of
\frac{1-y/x}{1+y/x}
as requested.
 
anuttarasammyak said:
\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}
where b is any positive number. So we get the formula of
\frac{1-y/x}{1+y/x}
as requested.
I do not seem to get what you did here...how did you arrive at this
\frac{y}{x}=\frac{log_na}{log_nc}=\frac{log_ba}{log_bc}
 
Between the middle and the right terms, there is a term
\frac{\log_ba/\log_bn}{\log_bc/\log_bn}
 
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