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UPSC NDA 2 2021 Mathematics Solved Paper

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Question : 40 of 120

Marks: +1, -0

If

\log_{10}2 \log_{2}10 + \log_{10}(10^{x}) = 2

, then what is the value of

x

?

0
1
$\log_{2}10$
log52

Solution:

Concept:

Use logarithm properties:

\log_{b}(b^{x}) = x

,

\log_{b}b = 1

, and the product of a log and its reciprocal equals

1

.

Explanation:

The given equation is

\log_{10}2 \cdot \log_{2}10 + \log_{10}(10^{x}) = 2

.

First, simplify

\log_{10}2 \cdot \log_{2}10

.

Using the change of base,

\log_{10}2 = \frac{\log 2}{\log 10}

and

\log_{2}10 = \frac{\log 10}{\log 2}

.

Their product is

\frac{\log 2}{\log 10} \times \frac{\log 10}{\log 2} = 1

.

Next, simplify

\log_{10}(10^{x})

using

\log_{b}(b^{x}) = x

, so

\log_{10}(10^{x}) = x

.

Substitute these into the equation:

1 + x = 2

.

Solve for

x

:

x = 2 - 1 = 1

.

Thus, the value of

x

is

1

.

Answer:

x = 1

(Option B).

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