Right angled triangle log If a,b and c(c is the hypotenuse) are sides of a right triangle then prove (log_(c+b)a)+(log_(c-b)a)=2( log_(c+b) a )*( log_(c-b)a) The bases are different so can't quite figure out what to do

Tiffany Page

Tiffany Page

Answered question

2022-11-03

Right angled triangle log
If a,b and c(c is the hypotenuse) are sides of a right triangle then prove
( log c + b a ) + ( log c b a ) = 2 ( log c + b a ) ( log c b a )
The bases are different so can't quite figure out what to do

Answer & Explanation

Justin Blake

Justin Blake

Beginner2022-11-04Added 11 answers

Since c 2 b 2 = a 2 , one has
log c + b ( a ) + log c b ( a ) = log a ( a ) log a ( c + b ) + log a ( a ) log a ( c b ) = log a ( c b ) + log a ( c + b ) log a ( c + b ) × log a ( c b ) = log a ( c 2 b 2 ) log a ( c + b ) × log a ( c b ) = log a ( a 2 ) log a ( c + b ) × log a ( c b ) = 2 log a ( c + b ) × log a ( c b ) = 2 log a ( a ) log a ( c + b ) log a ( a ) log a ( c b ) = 2 log c + b ( a ) × log c b ( a )
Dylan Benitez

Dylan Benitez

Beginner2022-11-05Added 2 answers

Recall that a 2 + b 2 = c 2 and log b a = ln a ln b (actually the base e is arbitrary, I use it for clarity). You are then asked to prove that
ln a ln ( c + b ) + ln a ln ( c b ) = 2 ln 2 a ln ( c + b ) ln ( c b ) ln ( c b ) + ln ( c + b ) = 2 ln a ln ( ( c b ) ( c + b ) ) = ln a 2 ln ( c 2 b 2 ) = ln a 2 c 2 b 2 = a 2 a 2 + b 2 = c 2

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