7 log16-15-5 log25-24-3 log81-80 is equal to -

The correct option is C log 27 log ( 16/15 )+5 log ( 25/24 )+3 log ( 81/80 ) ∵ a log x = log x^a ⇒ log (16/15)^7 + log (25/24)^5 + log ( 81/80)^3 ∵ log ...

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Byju's Answer

Standard X

Mathematics

Properties of Logarithms

7 log16/15+5 ...

Question

$7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})$ is equal to -

0.0

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log 2

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log 3

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Solution

The correct option is C log 2
$7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})$
$∵ a l o g x = l o g x^{a}$
$\Rightarrow l o g (\frac{16}{15})^{7} + l o g (\frac{25}{24})^{5} + l o g (\frac{81}{80})^{3}$
$∵ l o g x + l o g y = l o g (x \times y)$
$\Rightarrow l o g (\frac{16^{7}}{15^{7}} . \frac{25^{5}}{24^{5}} . \frac{81^{3}}{80^{3}})$
$= l o g 2$

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Similar questions

Q. The value of

7 log (\frac{16}{15}) + 5 log (\frac{25}{24}) + 3 log (\frac{81}{80})

is equal to:

Q. Show that :

7 log (\frac{16}{15}) + 5 log (\frac{25}{24}) + 3 log (\frac{81}{80}) = log 2

Q. Prove that

7 log \frac{16}{15} + 5 log \frac{25}{24} + 3 log \frac{81}{80} = log 2

Q. The value of

7 log (\frac{16}{15}) + 5 log (\frac{25}{24}) + 3 log (\frac{81}{80})

Q. The value of

[3 log (\frac{81}{80}) + 5 log (\frac{25}{24}) + 7 log (\frac{16}{15})]

is:

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7 log (1615)+5 log(2524)+3 log(8180) is equal to -

The correct option is C log 27 log (1615)+5 log(2524)+3 log(8180) ∵a log x=log xa ⇒log (1615)7+log (2524)5+log (8180)3 ∵log x+log y=log (x×y) ⇒log(167157.255245.813803) =log 2

$7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})$ is equal to -