Question

If $x^a=x^{\frac{b}{2}}{z}^{\frac{b}{2}}=z^c$, then prove that $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.

Answer 1

Here, $x^a=(xz{)}^{\frac{b}{2}}=z^c$

Now, taking log on both the sides:

$log(x{)}^a=log(xz{)}^{\frac{b}{2}}=log(z{)}^c$

$\Rightarrow alogx=\frac{b}{2}log\left(xz\right)=clogz$

$\Rightarrow alogx=\frac{b}{2}logx+\frac{b}{2}logz=clogz$

$\Rightarrow alogx=\frac{b}{2}logx+\frac{b}{2}logz$ and $\frac{b}{2}log$

$x+\frac{b}{2}logz=clogz$

$\Rightarrow (a-\frac{b}{2})logx=\frac{b}{2}logz$ and $\frac{b}{2}logx$ = $(c-\frac{b}{2})logz$

$\Rightarrow \frac{logx}{logz}=\frac{\frac{b}{2}}{(a-\frac{b}{2})}$ and \frac{log~x}{log~z} = \frac{\left(c - \frac{b}{2}\right)}{\frac{b}{2}}\)

$\Rightarrow \frac{\frac{b}{2}}{(a-\frac{b}{2})}=\frac{(c-\frac{b}{2})}{\frac{b}{2}}$

$\Rightarrow \frac{\frac{b}{2}}{(a-\frac{b}{2})}=\frac{(c-\frac{b}{2})}{\frac{b}{2}}$

$\Rightarrow \frac{b^2}{4}=ac-\frac{ab}{2}-\frac{bc}{2}+\frac{b^2}{4}$

$\Rightarrow 2ac=ab+bc$

$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

Thus, $\frac{1}{a},\frac{1}{b}$ and $\frac{1}{c}$ are in A.P.