solve this question and ill mark u as brainlist
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See the attachment........
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Heya!
To prove:
log |(√x²+1)+x| + log |(√x²+1)-x| = 0
Proof:
LHS
= log |(√x²+1)+x| + log |(√x²+1)-x|
= log | [(√x²+1)+x] [(√x²+1)-x] |
= log |(√x²+1)²-x²|
= log | x² + 1 -x²|
= log |1|
= log 1
= 0
Identities used:
• (x+y)(x-y) = x²-y²
• logA + logB = log (AB)
• log 1 = 0
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