Question

If \(\log_{10} 11 = a\) then \(\log_{10} \left(\frac{1}{110}\right)\) is equal to?

• A. \(-a\)
• B. \((1 + a)^{-1}\)
• C. \(\frac{1}{10a}\)
• D. \(-(a + 1)\)

Answer 1

The Correct Option is D

Given: \(\log_{10} 11 = a\)
We need to find \(\log_{10} \left(\frac{1}{110}\right)\).
We can express \(\frac{1}{110}\) as:
\[\frac{1}{110} = \frac{1}{11 \times 10} = \frac{1}{11} \times \frac{1}{10}\]
Using logarithm properties:
\[\log_{10} \left(\frac{1}{110}\right) = \log_{10} \left(\frac{1}{11 \times 10}\right) = \log_{10} \left(\frac{1}{11}\right) + \log_{10} \left(\frac{1{10}\right)
\]
We know:
\[\log_{10} \left(\frac{1}{x}\right) = -\log_{10} x\]
So:
\[\log_{10} \left(\frac{1}{11}\right) = -\log_{10} 11 = -a\]
And:
\[\log_{10} \left(\frac{1}{10}\right) = -\log_{10} 10 = -1\]
Adding these together:
\[\log_{10} \left(\frac{1}{110}\right) = -a + (-1) = -(a + 1)\]
Thus, the answer is:
D) \(-(a + 1)\)

Answer 2

Since \(\log_{10}11=a\), by the definition of a logarithm, \(11=10^{a}\). We can use this to write 110 as a power of 10 directly.

\[110 = 11 \times 10 = 10^{a}\times 10^{1} = 10^{a+1}\]
Therefore:
\[\frac{1}{110} = 10^{-(a+1)}\]
Taking \(\log_{10}\) of both sides:
\[\log_{10}\left(\frac{1}{110}\right) = -(a+1)\]

1. Option A \(-a\): This ignores the factor of 10, so it is incorrect.
2. Option B \((1+a)^{-1}\): This is a reciprocal, not a negative sum, so it is incorrect.
3. Option C \(\frac{1}{10a}\): This does not follow from the logarithm rules used, so it is incorrect.
4. Option D \(-(a+1)\): This matches the result obtained directly from the exponential form.

So \(\log_{10}\left(\frac{1}{110}\right) = -(a+1)\).

Hence, the correct answer is Option D: \(-(a+1)\).