Question

if 1/x^a=1/y^b=1/z^c and xz=y² then prove that a+c =2b​

Answer 1

Before we solve

In the question, we see equations for exponents. The equations may seem tough to solve. The first step to solve the question is to rewrite the equations and make them easy to solve. We rewrite the equations with the help of $k$ and exponent laws. So let's see how we can solve this problem.

Correct Question

Prove that $\dfrac{1}{a} +\dfrac{1}{c} =\dfrac{2}{b}$.

Solution

Step 1. Rewriting the Equations

Given

• $(\dfrac{1}{x} )^a=(\dfrac{1}{y} )^b=(\dfrac{1}{z} )^c$ ...(1)
• $xz=y^2$ ...(2)

In the first equation, let's add a parameter $k$

$\rightarrow (\dfrac{1}{x} )^a=(\dfrac{1}{y} )^b=(\dfrac{1}{z} )^c=k$

Then we write equations for $x,y,z$

• $(\dfrac{1}{x} )^a=k\rightarrow \dfrac{1}{x} =k^{\frac{1}{a}}$
• $(\dfrac{1}{y} )^b=k\rightarrow \dfrac{1}{y} =k^{\frac{1}{b}}$
• $(\dfrac{1}{z} )^c=k\rightarrow \dfrac{1}{z} =k^{\frac{1}{c}}$

Furthermore

• $\dfrac{1}{x} =k^{\frac{1}{a}}\rightarrow x=k^{-\frac{1}{a} }$ ...(1-1)
• $\dfrac{1}{y} =k^{\frac{1}{b}}\rightarrow y=k^{-\frac{1}{b} }$ ...(1-2)
• $\dfrac{1}{z} =k^{\frac{1}{c}}\rightarrow z=k^{-\frac{1}{c} }$ ...(1-3)

Step 2. Using the given Equation

From (2)

$\rightarrow xz=y^2$

$\rightarrow k^{-\frac{1}{a} -\frac{1}{c} }=k^{-\frac{2}{b} }$  ∵(1-1), (1-2), (1-3)

$\rightarrow -\dfrac{1}{a} -\dfrac{1}{c} =-\dfrac{2}{b}$

$\rightarrow \dfrac{1}{a} +\dfrac{1}{c} =\dfrac{2}{b}$

Hence proven.