If a^2+b^2=c^2, 1/log with base (b+c) of a+1/log with base(c-b) of a is
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Given a^2+b^2=c^2
a^2=c^2-b^2
a^2=(c+b)(c-b)-----(1)
[1/(log base xof y) = log base y ofx ]
Now
Log base a of (b+c) + log base a of(c-b)
=log base a of [(b+c)(b-c)]
=log base a of a^2 since from (1)
=2log base a of a
=2
a^2=c^2-b^2
a^2=(c+b)(c-b)-----(1)
[1/(log base xof y) = log base y ofx ]
Now
Log base a of (b+c) + log base a of(c-b)
=log base a of [(b+c)(b-c)]
=log base a of a^2 since from (1)
=2log base a of a
=2
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