1.

Given `a^2+b^2=c^2& a .0 ; b >0; c >0, c-b!=1, c+b!=1, `prove that: `(log)_("c"+"b")a+(log)_("c"-"b")a=2(log)_("c"+"b")adot(log)_("c"-"b")a`A. 1B. 2C. `-1`D. `-2`

Answer» Correct Answer - B
We have,
`(1)/("log"_(c+a)b) + (1)/("log"_(c-a)b)`
` = "log"b (c+a) + "log"_(b) (c-a)`
` = "log"_(b) (c^(2)-a^(2)) = "log"_(b)b^(2) " " [because c^(2) -a^(2) = b^(2)]`
`2 "log"_(b) b= 2`


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