7 log16-15-5 log25-25-3 log81-80 is equal to

The correct option is C log 2Given expression =7 log [ 2^4/3× 5 ]+5 log [ 5^2/3× 2^3 ]+3log [ 3^4/5× 2^4 ] =7 [4 log2 log3 log5]+5[2log5 log3 3log2]+3[4log3 lo ...

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Standard IX

Mathematics

Standard Logarithm

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

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log 2

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log 3

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Solution

The correct option is B log 2
Given expression
$= 7 l o g [\frac{2^{4}}{3 \times 5}] + 5 l o g [\frac{5^{2}}{3 \times 2^{3}}] + 3 l o g [\frac{3^{4}}{5 \times 2^{4}}]$
$= 7 [4 l o g 2 - l o g 3 - l o g 5] + 5 [2 l o g 5 - l o g 3 - 3 l o g 2] + 3 [4 l o g 3 - l o g 5 - 4 l o g 2]$
$= l o g 2$

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Similar questions

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

is equal to

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})

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

The correct option is B log 2Given expression =7log[243×5]+5log[523×23]+3log[345×24] =7[4log2−log3−log5]+5[2log5−log3−3log2]+3[4log3−log5−4log2] =log2

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

The correct option is B log 2
Given expression
$= 7 l o g [\frac{2^{4}}{3 \times 5}] + 5 l o g [\frac{5^{2}}{3 \times 2^{3}}] + 3 l o g [\frac{3^{4}}{5 \times 2^{4}}]$
$= 7 [4 l o g 2 - l o g 3 - l o g 5] + 5 [2 l o g 5 - l o g 3 - 3 l o g 2] + 3 [4 l o g 3 - l o g 5 - 4 l o g 2]$
$= l o g 2$