Honors Algebra 2 B Print UNIT 10 LESSON 2 Honors Algebra 2 B Semester Review and Exam Honors Algebra 2 B Semester Exam Mark as Complete page 1 of 1 Algebra 2B Semester Exam 5 of 40 COURSE TOOLS The expression 1og, 26 was evaluated using the change of base formula with base e as 2.966. What is the value of the expression log 26 when evaluated using the change of base formula with base 5? (1 point) 2.024 Next

1 answer

To evaluate the expression \(\log_5 26\) using the change of base formula, we can apply the formula as follows:

\[ \log_b a = \frac{\log_k a}{\log_k b} \]

where \(k\) can be any base. In this case, we will use base \(e\) (natural logarithm, denoted as \(\ln\)) for consistency since we already have a value for \(\log_{10} 26\).

Given:

  • \(\log_e 26 \approx 2.966\)

We need to find \(\log_5 26\):

\[ \log_5 26 = \frac{\log_e 26}{\log_e 5} \]

Now, you would need to find \(\log_e 5\). If not given, we can find it using common logarithms:

Using the change of base formula, we can find approximately:

\[ \log_e 5 = \frac{\log_{10} 5}{\log_{10} e} \]

Where typically:

  • \(\log_{10} 5 \approx 0.699\)
  • \(\log_{10} e \approx 0.434\)

So:

\[ \log_e 5 \approx \frac{0.699}{0.434} \approx 1.61 \]

Now substituting back into the equation for \(\log_5 26\):

\[ \log_5 26 = \frac{2.966}{1.61} \approx 1.84 \]

Therefore, \(\log_5 26\) is approximately 1.84.

If you need a more precise value, you'd calculate the exact value of \(\log_e 5\) using a scientific calculator or find it directly.