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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 39 of 120

Marks: +1, -0

IF x + \log_{15} (1 + 3^{x}) = x \log_{15} 5 + \log_{15} 12, \text{ where } x \text{ is an integer, then what is } x \text{ equal to?}

$-3$
2
1
3

Solution:

Concept:

Use logarithmic properties to combine and equate arguments.

Explanation:

Given:

x + \log_{15}(1 + 3^x) = x \log_{15} 5 + \log_{15} 12

.

Rewrite

x

as

\log_{15} 15^x

.

Thus:

\log_{15} 15^x + \log_{15}(1 + 3^x) = \log_{15} 5^x + \log_{15} 12

.

Combine logs:

\log_{15}[15^x(1 + 3^x)] = \log_{15}[5^x \cdot 12]

.

Equate arguments:

15^x(1 + 3^x) = 5^x \cdot 12

.

Divide both sides by

5^x

:

\left(\frac{15}{5}\right)^x (1 + 3^x) = 12

→

3^x(1 + 3^x) = 12

.

Factor

12 = 3 \cdot 4

and note

4 = 1 + 3

. So

3^x(1 + 3^x) = 3(1 + 3)

.

This suggests

x = 1

works:

3^1(1 + 3^1) = 3 \cdot 4 = 12

.

Alternatively, let

y = 3^x

. Then

y(1+y) = 12

→

y^2 + y - 12 = 0

→

(y+4)(y-3)=0

.

Thus

y = 3

or

y = -4

(invalid). So

3^x = 3

→

x = 1

.

Answer:

x = 1

(Option C).

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