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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: