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UPSC NDA I 2026 Mathematics Paper

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Question : 13 of 120

Marks: +1, -0

If

1-\log _{10} 2=\log _{10}\left(5^{x}+4^{x}+3^{x}+2^{x}+1\right)

, then what is a value of

x

?

10
5
1
0

Solution:

Concept:

Use

\log_{10}10 = 1

and

\log A - \log B = \log \frac{A}{B}

to simplify the left side.

Explanation:

Given:

1-\log_{10}2=\log_{10}(5^x+4^x+3^x+2^x+1)

.

Since

1=\log_{10}10

,

\log_{10}10-\log_{10}2=\log_{10}\left(\frac{10}{2}\right)=\log_{10}5

.

So the equation becomes

\log_{10}5=\log_{10}(5^x+4^x+3^x+2^x+1)

.

Hence,

5^x+4^x+3^x+2^x+1=5

.

Check

x=0

:

5^0+4^0+3^0+2^0+1=1+1+1+1+1=5

.

So

x=0

.

Answer:

x=0

, which is option D.

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