Type: Proving, Topic: Logarithmic Equation
Question: If [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH] =1, prove [MATH]\frac{1}{\log_{x+y} a } [/MATH] = [MATH]\frac{1}{\log_x a }[/MATH] + [MATH]\frac{1}{\log_x y }[/MATH]
My attempt: trying to prove from the right side of the equation.
[MATH]\frac{1}{\log_x a }[/MATH] + [MATH]\frac{1}{\log_x y }[/MATH]
=logaxlogaa1+logyxlogyy1
=logax+logay
=logaxy
=logx+yalogx+yxy
=logx+yalogx+yx+logx+yy
=logx+yalogxx+ylogxx+logyx+ylogyy
=logx+yalogxx+y1+logyx+y1
Then I was stuck, having no idea to continue solving it...
My idea is trying to make the numerator logxx+y1+logyx+y1 to [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH], so that I could make it =1, as the question has given [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH] =1, but I can't think of any other ways to make the numerator to [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH].
Question: If [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH] =1, prove [MATH]\frac{1}{\log_{x+y} a } [/MATH] = [MATH]\frac{1}{\log_x a }[/MATH] + [MATH]\frac{1}{\log_x y }[/MATH]
My attempt: trying to prove from the right side of the equation.
[MATH]\frac{1}{\log_x a }[/MATH] + [MATH]\frac{1}{\log_x y }[/MATH]
=logaxlogaa1+logyxlogyy1
=logax+logay
=logaxy
=logx+yalogx+yxy
=logx+yalogx+yx+logx+yy
=logx+yalogxx+ylogxx+logyx+ylogyy
=logx+yalogxx+y1+logyx+y1
Then I was stuck, having no idea to continue solving it...
My idea is trying to make the numerator logxx+y1+logyx+y1 to [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH], so that I could make it =1, as the question has given [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH] =1, but I can't think of any other ways to make the numerator to [MATH]\frac{1}{x} [/MATH] + [MATH]\frac{1}{y} [/MATH].