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Mathematics

Write the logarithmic equation for :

x=ababa+bx = ab \sqrt{\dfrac{a - b}{a + b}}

Logarithms

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Answer

Given,

x=ababa+bTaking log on Both sides,log x=log (ababa+b)log x=log ab+log aba+blog x=log a+log b+log (aba+b)12log x=log a+log b+12log aba+blog x=log a+log b+12[log (ab)log (a+b)]\Rightarrow x = ab \sqrt{\dfrac{a - b}{a + b}} \\[1em] \text{Taking log on Both sides,} \\[1em] \Rightarrow \log \space x = \log \space \Big(ab \sqrt{\dfrac{a - b}{a + b}}\Big) \\[1em] \Rightarrow \log \space x = \log \space ab + \log \space\sqrt{\dfrac{a - b}{a + b}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \log \space \Big({\dfrac{a - b}{a + b}}\Big)^{\dfrac{1}{2}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \dfrac{1}{2} \log \space {\dfrac{a - b}{a + b}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \dfrac{1}{2} [\log \space ({a - b}) - \log \space ({a + b})] \\[1em]

Hence, logarithmic equation is

log x=log a+log b+12[log (ab)log (a+b)]\log \space x = \log \space a + \log \space b + \dfrac{1}{2} [\log \space ({a - b}) - \log \space ({a + b})]

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