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Sum of N Terms of an AP

log 2+16 log1...

Question

$l o g 2 + 16 l o g \frac{16}{15} + 12 l o g \frac{25}{24} + 7 l o g \frac{81}{80}$

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Solution

$l o g 2 + 16 l o g (\frac{16}{15}) + 12 l o g (\frac{25}{24}) 7 l o g (\frac{81}{80})$
$= l o g 2 + 16 [l o g 16 - l o g 15] + 12 [l o g 25 - l o g 24] + 7 [l o g 81 - l o g 80] [l ∵ l o g (\frac{a}{b}) = l o g a - l o g b]$
$= l o g 2 + 16 [l o g 2^{4} - l o g (3 \times 5)] + 12 [l o g 5^{2} - l o g (8 \times 3)] [∵ l o g (a)^{m} = m l o g a]$
$= l o g 2 + 64 l o g 2 - 16 l o g 3 - 16 l o g 5 + 24 l o g 5 - 36 l o g 2 - 12 l o g 3 + 28 l o g 3 - 7 l o g 5 - 28 l o g 2$
$= l o g 2 + l o g 5$

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Similar questions

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

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

\frac{16}{15}

\frac{25}{24}

\frac{81}{80}

Q. Prove that :

4 log \frac{24}{25} - 16 log \frac{9}{10} + 7 log \frac{81}{80} = log 5

Q. The value of,

log102+16log10(16/15)+12log10(25/24)+7log10(81/80)log10⁡2+16log10⁡(16/15)+12log10⁡(25/24)+7log10⁡(81/80) is

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