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Let \(F = \log \dfrac{1 + x}{1 - x}\)

 

Express \(G = \log \left( \dfrac{1 + \dfrac{3x + x^3}{1 + 3x^2}}{1 - \dfrac{3x + x^3}{1 + 3x^2}} \right)\) in terms of F.

 May 14, 2020

Best Answer 

 #1
avatar+8945 
+1

\(G\ =\ \log \left( \dfrac{1 + \dfrac{3x + x^3}{1 + 3x^2}}{1 - \dfrac{3x + x^3}{1 + 3x^2}} \right)\\~\\~\\ G\ =\ \log \left( \dfrac{\dfrac{1+3x^2}{1+3x^2} + \dfrac{3x + x^3}{1 + 3x^2}}{\dfrac{1+3x^2}{1+3x^2} - \dfrac{3x + x^3}{1 + 3x^2}} \right)\\~\\~\\ G\ =\ \log \left( \dfrac{\dfrac{x^3+3x^2+3x+1}{1+3x^2} }{\dfrac{-x^3+3x^2-3x+1}{1+3x^2} } \right)\\~\\~\\ G\ =\ \log \left( \dfrac{x^3+3x^2+3x+1 }{-x^3+3x^2-3x+1 } \right)\\~\\~\\ G\ =\ \log \left( \dfrac{(1+x)^3 }{(1-x)^3 } \right)\\~\\~\\ G\ =\ \log \left( \left( \dfrac{1+x }{1-x } \right)^3 \right)\\~\\~\\ G\ =\ 3 \log \left( \dfrac{1+x }{1-x } \right)\\~\\~\\ G\ =\ 3 F \)  .

 May 14, 2020
 #1
avatar+8945 
+1
Best Answer

\(G\ =\ \log \left( \dfrac{1 + \dfrac{3x + x^3}{1 + 3x^2}}{1 - \dfrac{3x + x^3}{1 + 3x^2}} \right)\\~\\~\\ G\ =\ \log \left( \dfrac{\dfrac{1+3x^2}{1+3x^2} + \dfrac{3x + x^3}{1 + 3x^2}}{\dfrac{1+3x^2}{1+3x^2} - \dfrac{3x + x^3}{1 + 3x^2}} \right)\\~\\~\\ G\ =\ \log \left( \dfrac{\dfrac{x^3+3x^2+3x+1}{1+3x^2} }{\dfrac{-x^3+3x^2-3x+1}{1+3x^2} } \right)\\~\\~\\ G\ =\ \log \left( \dfrac{x^3+3x^2+3x+1 }{-x^3+3x^2-3x+1 } \right)\\~\\~\\ G\ =\ \log \left( \dfrac{(1+x)^3 }{(1-x)^3 } \right)\\~\\~\\ G\ =\ \log \left( \left( \dfrac{1+x }{1-x } \right)^3 \right)\\~\\~\\ G\ =\ 3 \log \left( \dfrac{1+x }{1-x } \right)\\~\\~\\ G\ =\ 3 F \)  .

hectictar May 14, 2020
 #2
avatar+111360 
+1

Impressive, hectictar   !!!!

 

 

cool cool cool

CPhill  May 14, 2020

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