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Logarithms

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Question : 68 of 85

Marks: +1, -0

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

is equal to

1
$\log 2$
$\log 3$

Solution:

Concept:

Use logarithm properties and prime factorization to simplify the given expression.

Explanation:

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

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

= \log\left(\frac{16^7 \cdot 25^5 \cdot 81^3}{15^7 \cdot 24^5 \cdot 80^3}\right)

Express each number as product of primes:

16=2^4

,

25=5^2

,

81=3^4

;

15=3\cdot5

,

24=2^3\cdot3

,

80=2^4\cdot5

.

Then numerator:

2^{28} \cdot 5^{10} \cdot 3^{12}

Denominator:

(3^7\cdot5^7) \cdot (2^{15}\cdot3^5) \cdot (2^{12}\cdot5^3) = 2^{27} \cdot 3^{12} \cdot 5^{10}

Thus expression

= \log\left(\frac{2^{28} \cdot 5^{10} \cdot 3^{12}}{2^{27} \cdot 3^{12} \cdot 5^{10}}\right) = \log(2)

.

Answer:

\log 2

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