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UPSC NDA I and II 2020 Math Solved Paper

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Question : 36 of 120

Marks: +1, -0

What is the value of

\frac{1}{10}log_{5}1024-log_{5}10+\frac{1}{5}log_{5}3125

?

0
1
2
3

Solution:

Concept:

The solution uses the logarithm properties:

\log_b(m^n)=n\log_b m

and

\log_b(mn)=\log_b m+\log_b n

.

Explanation:

Start with the given expression:

\frac{1}{10}\log_5{1024}-\log_5{10}+\frac{1}{5}\log_5{3125}

.

Rewrite each number as a power of a small base:

1024 = 2^{10}

,

10 = 2\times 5

, and

3125 = 5^5

.

Substitute:

\frac{1}{10}\log_5{2^{10}} - \log_5(2\times 5) + \frac{1}{5}\log_5{5^5}

.

Apply the power property:

\frac{1}{10}\cdot 10\log_5 2 - \log_5(2\times 5) + \frac{1}{5}\cdot 5\log_5 5 = \log_5 2 - \log_5(2\times 5) + \log_5 5

.

Now use the product property:

\log_5(2\times 5) = \log_5 2 + \log_5 5

.

Thus, expression becomes:

\log_5 2 - (\log_5 2 + \log_5 5) + \log_5 5 = \log_5 2 - \log_5 2 - \log_5 5 + \log_5 5 = 0

.

Answer:

0

(Option A).

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