Question

If ay = by = cz and b2=ac, prove that y= (2xz/x+y).​

Answer 1

$\textsf{\large{\underline{Correct Question}:}}$

• If aˣ = bʸ = cᶻ and b² = ac, prove that y = 2xz/(x + z)

$\textsf{\large{\underline{Solution}:}}$

Let us assume that:

$\rm: \longmapsto {a}^{x} = {b}^{y} = {c}^{z} = k$

Therefore, we can say that:

$\rm: \longmapsto {a}^{x}= k \: \: or \: \: a = {k}^{^{1}/_{x} }$

$\rm: \longmapsto {b}^{y}= k \: \: or \: \: b= {k}^{^{1}/_{y} }$

$\rm: \longmapsto {c}^{z}= k \: \: or \: \: c= {k}^{^{1}/_{z} }$

Now, it's given that:

$\rm: \longmapsto {b}^{2} = ac$

$\rm: \longmapsto {({k}^{^{1}/_{y}})}^{2} ={k}^{^{1}/_{x}} \times {k}^{^{1}/_{z}}$

$\rm: \longmapsto {k}^{^{2}/_{y}} ={k}^{^{1}/_{x} +^{1}/_{z} }$

Comparing base, we get:

$\rm: \longmapsto \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}$

$\rm: \longmapsto \dfrac{2}{y} = \dfrac{x + z}{xz}$

$\rm: \longmapsto \dfrac{1}{y} = \dfrac{x + z}{2xz}$

$\rm: \longmapsto y = \dfrac{2xz}{x + z}$

Hence, proved.

$\textsf{\large{\underline{More To Know}:}}$

Laws Of Exponents: If a, b are positive real numbers and m, n are rational numbers, then the following results hold.

$\rm 1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm 2. \: \: ({a}^{m})^{n} = {a}^{mn}$

$\rm 3. \: \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\rm4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\rm5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\rm6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\rm7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\rm8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$