## A problem of log, GP and HP…

Question: If $a^{x}=b^{y}=c^{z}$ and $b^{2}=ac$, pyrove that: $y = \frac{2xz}{x+z}$

Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;

That is, to prove that $y=\frac{2xz}{x+z}$ is the same as the proof for : $\frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}$

Now, it is given that $a^{x} = b^{y} = c^{z}$ —– I

and $b^{2}=ac$ —– II
Let $a^{x} = b^{y}=c^{z}=N$ say. By definition of logarithm, $x = \log_{a}{N}$; $y=\log_{b}{N}$; $z=\log_{c}{N}$ $\frac{1}{x} = \frac{1}{\log_{a}{N}}$; $\frac{1}{y} = \frac{1}{\log_{b}{N}}$; $\frac{1}{z} = \frac{1}{\log_{a}{N}}$.

Now let us see what happens to the following two algebraic entities, namely, $\frac{1}{y} - \frac{1}{x}$ and $\frac{1}{z} - \frac{1}{y}$;

Now, $\frac{1}{y} - \frac{1}{x} = \frac{1}{\log_{b}{N}} - \frac{1}{\log_{a}{N}} = \frac{\log_{b}{b}}{\log_{b}{N}} - \frac{\log_{a}{a}}{\log_{a}{N}} = \log_{N}{b} - \log_{N}{a} = \log_{N}{(\frac{b}{a})}$…call this III

Now, $\frac{1}{z} - \frac{1}{y} = \frac{1}{\log_{c}{N}} - \frac{1}{\log_{b}{N}} = \frac{\log_{c}{c}}{\log_{c}{}N} -\frac{\log_{b}{b}}{\log_{b}{N}}= \log_{N}{c}-\log_{N}{b}$

Hence, $\frac{1}{z} - \frac{1}{y}=\log_{N}{c/b}$….equation IV

but it is also given that $b^{2}=ac$…see equation II

Hence, $\frac{b}{a} = \frac{c}{b}$

Take log of above both sides w.r.t. base N:

So, above is equivalent to $\log_{N}{b/a} = \log_{N}{c/b}$

But now see relations III and IV:

Hence, $\frac{1}{y} -\frac{1}{x} = \frac{1}{x} - \frac{1}{y}$

Hence, $\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}$

Hence, $y= \frac{2xz}{x+z}$ as desired.

Regards,

Nalin Pithwa

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