Type: Proving, Topic: Logarithmic Equation
Question: If [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], prove [MATH]\frac{1}{\log_{x+y} a } [/MATH] [math]=[/math] [MATH]\frac{1}{\log_x a }[/MATH] [math]+[/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]+[/math] [MATH]\frac{1}{\log_x y }[/MATH]
[math]= \frac{1}{\frac{\log_a a }{\log_a x }} + \frac{1}{\frac{\log_y y }{\log_y x }}[/math]
[math]= \log_a x + \log_a y[/math]
[math]= \log_a {xy}[/math]
[math]= \frac{\log_{x+y} xy }{\log_{x+y} a }[/math]
[math]= \frac{\log_{x+y} x + \log_{x+y} y}{\log_{x+y} a }[/math]
[math]= \frac{\frac{\log_x x}{\log_x {x+y}} + \frac{log_y y}{\log_y {x+y}}}{\log_{x+y} a }[/math]
[math]= \frac{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}{\log_{x+y} a }[/math]
Then I was stuck, having no idea to continue solving it...
My idea is trying to make the numerator [math]{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}[/math] to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH], so that I could make it [math]= 1[/math], as the question has given [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], but I can't think of any other ways to make the numerator to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH].
Question: If [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], prove [MATH]\frac{1}{\log_{x+y} a } [/MATH] [math]=[/math] [MATH]\frac{1}{\log_x a }[/MATH] [math]+[/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]+[/math] [MATH]\frac{1}{\log_x y }[/MATH]
[math]= \frac{1}{\frac{\log_a a }{\log_a x }} + \frac{1}{\frac{\log_y y }{\log_y x }}[/math]
[math]= \log_a x + \log_a y[/math]
[math]= \log_a {xy}[/math]
[math]= \frac{\log_{x+y} xy }{\log_{x+y} a }[/math]
[math]= \frac{\log_{x+y} x + \log_{x+y} y}{\log_{x+y} a }[/math]
[math]= \frac{\frac{\log_x x}{\log_x {x+y}} + \frac{log_y y}{\log_y {x+y}}}{\log_{x+y} a }[/math]
[math]= \frac{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}{\log_{x+y} a }[/math]
Then I was stuck, having no idea to continue solving it...
My idea is trying to make the numerator [math]{\frac{1}{\log_x {x+y}} + \frac{1}{\log_y {x+y}}}[/math] to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH], so that I could make it [math]= 1[/math], as the question has given [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH] [math]= 1[/math], but I can't think of any other ways to make the numerator to [MATH]\frac{1}{x} [/MATH] [math]+[/math] [MATH]\frac{1}{y} [/MATH].