Question

Zahra is finding an approximation to four decimal places of log11. What is the upper bound approximation?
A. 1.04140


D. 1.04139

Answer 1

To approximate \(\log_{11}\), we can use the change of base formula:

\[
\log_{11} = \frac{\log_{10}}{\log_{10}(11)}
\]

Using a calculator, we can find the value of \(\log_{10}(11)\):

\[
\log_{10}(11) \approx 1.04139
\]

Thus, we can write:

\[
\log_{11} \approx \frac{\log_{10}(n)}{1.04139}
\]

Since we are looking for an upper bound approximation between the provided options, let's analyze the values:

- A. \(1.04140\)
- D. \(1.04139\)

Since the calculated \(\log_{10}(11)\) is approximately \(1.04139\), and we want to approximate \(\log_{11}\) to four decimal places, the upper bound options suggest rounding we can assume the actual value would be nearest to that.

Considering that \(1.04140\) is greater than \(1.04139\) and would signify an upper bound for \(\log_{11}\) within four decimal places, while \(1.04139\) would be an accurate representation, the upper bound approximation of \(\log_{11}\) is:

\[
\text{Upper bound approximation: } A. 1.04140
\]