Iflogab+logba=log(a+b),then−Iflog⁡ab+log⁡ba=log⁡(a+b),then−{\text{If}}\,\,\,\log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right),\,\,{\text{then}} - " /> Iflogab+logba=log(a+b),then−Iflog⁡ab+log⁡ba=log⁡(a+b),then−{\text{If}}\,\,\,\log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right),\,\,{\text{then}} - " /> Iflogab+logba=log(a+b),then−Iflog⁡ab+log⁡ba=log⁡(a+b),then−{\text{If}}\,\,\,\log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right),\,\,{\text{then}} - " />

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$If \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b), then -$ ${\text{If}}\,\,\,\log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right),\,\,{\text{then}} -$

1a + b = 1
2a - b = 1
3a = b
4a² - b² = 1

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b) \\ \Rightarrow \log (a + b) = \log (\frac{a}{b} \times \frac{b}{a}) \\ \Rightarrow \log (a + b) = \log 1 \\ S o, a + b = 1 \end{aligned}

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", "text": "

If \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b), then -

${\text{If}}\,\,\,\log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right),\,\,{\text{then}} -$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \log \frac{a}{b} + \log \frac{b}{a} = \log (a + b) \\ \Rightarrow \log (a + b) = \log (\frac{a}{b} \times \frac{b}{a}) \\ \Rightarrow \log (a + b) = \log 1 \\ S o, a + b = 1 \end{aligned}

$\eqalign{ & \log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log \left( {\frac{a}{b} \times \frac{b}{a}} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log 1 \cr & So,\,\,\,a + b = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: