If(ab)x−1=(ba)x−3,thenthevalueofxisIf(ab)x−1=(ba)x−3,thenthevalueofxis\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr} " /> If(ab)x−1=(ba)x−3,thenthevalueofxisIf(ab)x−1=(ba)x−3,thenthevalueofxis\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr} " /> If(ab)x−1=(ba)x−3,thenthevalueofxisIf(ab)x−1=(ba)x−3,thenthevalueofxis\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr} " />

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$\begin{aligned} If {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3}, \\ then the value of x is \end{aligned}$ $\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3} \\ \Rightarrow {(\frac{a}{b})}^{x - 1} = {(\frac{a}{b})}^{- (x - 3)} = {(\frac{a}{b})}^{(3 - x)} \\ \Rightarrow x - 1 = 3 - x \\ \Rightarrow 2 x = 4 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr}$

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", "text": "

\begin{aligned} If {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3}, \\ then the value of x is \end{aligned}

$\eqalign{ & {\text{If}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}}, \cr & {\text{then}}\,{\text{the}}\,{\text{value}}\,{\text{of}}\,x\,{\text{is}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3} \\ \Rightarrow {(\frac{a}{b})}^{x - 1} = {(\frac{a}{b})}^{- (x - 3)} = {(\frac{a}{b})}^{(3 - x)} \\ \Rightarrow x - 1 = 3 - x \\ \Rightarrow 2 x = 4 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Given {(\frac{a}{b})}^{x - 1} = {(\frac{b}{a})}^{x - 3} \\ \Rightarrow {(\frac{a}{b})}^{x - 1} = {(\frac{a}{b})}^{- (x - 3)} = {(\frac{a}{b})}^{(3 - x)} \\ \Rightarrow x - 1 = 3 - x \\ \Rightarrow 2 x = 4 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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