MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Property 3

If log a-b/2...

Question

If $l o g \frac{a - b}{2} = \frac{1}{2} (l o g a + l o g b)$ , show that:
$a^{2} + b^{2} = 6 a b$

Open in App

Solution

log $open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =(logab)
log $open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =log
$open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets$ =
$open square brackets fraction numerator a minus b over denominator 2 end fraction close square brackets squared$ =ab
=ab
=4ab
-2ab=4ab
=6ab

Suggest Corrections

thumbs-up

6

Similar questions

log (\frac{a - b}{2}) = \frac{1}{2} (log a + log b)

, then the value of

a^{2} + b^{2} - 6 a b

a^{2} + b^{2} = 23 a b

log \frac{a + b}{5} = \frac{1}{2} (log a + log b)

a^{2} + b^{2} = 3 a b,

log (\frac{a + b}{\sqrt{5}}) = \frac{1}{2} (log a + log b)

a^{2} + b^{2} = 7 a b

, then prove that

log (\frac{a + b}{3}) = \frac{1}{2} (log a + log b)

a^{2} - 12 a b + 4 b^{2} = 0

log (a + 2 b) = \frac{1}{2} (log a + log b) + 2 log 2

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Laws of Logarithm with Use

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon