If log a-b/2=1/2(log a+log b), show that: a^2+b^2 = 6ab
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log ( a - b )/2 = 1/2 ( log a + log b )
2 log ( a - b ) /2 = log a + log b
log [ ( a - b )/2 ] ² = log ab
( a - b )²/2² = ab
( a - b )² = 4ab
a² + b² - 2ab = 4ab
a² + b² = 4ab + 2ab
a² + b² = 6ab
I hope this helps you
:)
2 log ( a - b ) /2 = log a + log b
log [ ( a - b )/2 ] ² = log ab
( a - b )²/2² = ab
( a - b )² = 4ab
a² + b² - 2ab = 4ab
a² + b² = 4ab + 2ab
a² + b² = 6ab
I hope this helps you
:)
rawatrishabh03:
thanks bro it was really helpful
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