Question

If we have an expression as ${{x}^{a}}={{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}={{z}^{c}}$, then prove that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in A.P

Answer 1

Hint: In this problem, we are given three equalities with three terms. We will consider 2 terms at a first and since we need to deal with the powers, apply ln (${{\log }_{e}}$) on both sides. Then we will try to find the relation between the powers. Then we will take the next equality and repeat the same operation and try to find the relation between the powers. Then we will try to prove that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$are in arithmetic progression (AP).

Complete step-by-step solution:
Thus, to prove that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ to be in AP, we must prove the following relation.
$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$
Now, it is given to us that ${{x}^{a}}={{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}={{z}^{c}}$.
We must first consider the first equality, i.e. ${{x}^{a}}={{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}$.
We will apply ln on both sides.
 \[\Rightarrow \ln {{x}^{a}}=\ln {{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}\]
But we know that ln(ab) = ln(a) + ln(b).
\[\begin{align}
  & \Rightarrow \ln {{x}^{a}}=\ln {{x}^{\dfrac{b}{2}}}+\ln {{z}^{\dfrac{b}{2}}} \\
 & \Rightarrow a\ln x=\dfrac{b}{2}\ln x+\dfrac{b}{2}\ln z \\
 & \Rightarrow \dfrac{\left( a-\dfrac{b}{2} \right)}{\dfrac{b}{2}}\ln x=\ln z \\
 & \Rightarrow \dfrac{a-\dfrac{b}{2}}{\dfrac{b}{2}}=\dfrac{\ln z}{\ln x} \\
\end{align}\]
But we know that $\dfrac{\ln a}{\ln b}={{\log }_{b}}a$.
$\Rightarrow \dfrac{a-\dfrac{b}{2}}{\dfrac{b}{2}}={{\log }_{x}}z......\left( 1 \right)$
We shall now consider the first equality, i.e. ${{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}={{z}^{c}}$.
We will apply ln on both sides and follow the same steps we followed in the first equality.
 \[\begin{align}
  & \Rightarrow \ln {{x}^{\dfrac{b}{2}}}{{z}^{\dfrac{b}{2}}}=\ln {{z}^{c}} \\
 & \Rightarrow \ln {{z}^{c}}=\ln {{x}^{\dfrac{b}{2}}}+\ln {{z}^{\dfrac{b}{2}}} \\
 & \Rightarrow c\ln z=\dfrac{b}{2}\ln x+\dfrac{b}{2}\ln z \\
 & \Rightarrow \left( c-\dfrac{b}{2} \right)\ln z=\dfrac{b}{2}\ln x \\
 & \Rightarrow \dfrac{c-\dfrac{b}{2}}{\dfrac{b}{2}}=\dfrac{\ln x}{\ln z} \\
 & \Rightarrow \dfrac{c-\dfrac{b}{2}}{\dfrac{b}{2}}={{\log }_{z}}x \\
\end{align}\]
Now, if $\dfrac{\ln a}{\ln b}={{\log }_{b}}a$ then $\dfrac{\ln b}{\ln a}={{\log }_{a}}b$ and thus ${{\log }_{b}}a=\dfrac{1}{{{\log }_{a}}b}$.
$\begin{align}
  & \Rightarrow \dfrac{c-\dfrac{b}{2}}{\dfrac{b}{2}}=\dfrac{1}{{{\log }_{x}}z} \\
 & \Rightarrow \dfrac{\dfrac{b}{2}}{c-\dfrac{b}{2}}={{\log }_{z}}x......\left( 2 \right) \\
\end{align}$
Therefore, we can see that the right hand side of (1) and (2) are equal. Thus, left hand side must also be equal.
\[\Rightarrow \dfrac{a-\dfrac{b}{2}}{\dfrac{b}{2}}=\dfrac{\dfrac{b}{2}}{c-\dfrac{b}{2}}\]
Now, we shall simplify the above equation and try to reduce it to $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$.
\[\begin{align}
  & \Rightarrow \left( a-\dfrac{b}{2} \right)\left( c-\dfrac{b}{2} \right)={{\left( \dfrac{b}{2} \right)}^{2}} \\
 & \Rightarrow ac-\dfrac{ab}{2}-\dfrac{bc}{2}+{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{b}{2} \right)}^{2}} \\
\end{align}\]
\[{{\left( \dfrac{b}{2} \right)}^{2}}\]cancels out from both sides.
 \[\begin{align}
  & \Rightarrow ac-\dfrac{ab}{2}-\dfrac{bc}{2}=0 \\
 & \Rightarrow ac=\dfrac{ab}{2}+\dfrac{bc}{2} \\
 & \Rightarrow 1=\dfrac{b}{2}\left( \dfrac{1}{c}+\dfrac{1}{a} \right) \\
 & \Rightarrow \dfrac{2}{b}=\dfrac{1}{c}+\dfrac{1}{a}......\left( 3 \right) \\
\end{align}\]
Thus, from (3), we can say that $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{1}{c}$ are in A.P.

Note: For three terms to be in arithmetic progression, twice the middle term must be equal to the sum of the other two terms. Students must know about the operations of logarithm whenever it comes to dealing with the powers.

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