Question

log4 512
log8 200
log6 50
log5 100
log7 965
log4 1256
log6 7776

Answer 1

To convert these logarithms into a more manageable form, we can use the change of base formula, which states:

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

where \( k \) can be any positive number. A common choice for \( k \) is 10 or \( e \) (natural logarithm). Here, we will use base 10 logarithms for simplicity.

Let's calculate the logarithms you've provided:

1. **log₄(512)**:
\[
\log_4(512) = \frac{\log_{10}(512)}{\log_{10}(4)} \approx \frac{2.709}{0.602} \approx 4.50
\]

2. **log₈(200)**:
\[
\log_8(200) = \frac{\log_{10}(200)}{\log_{10}(8)} \approx \frac{2.301}{0.903} \approx 2.55
\]

3. **log₆(50)**:
\[
\log_6(50) = \frac{\log_{10}(50)}{\log_{10}(6)} \approx \frac{1.699}{0.778} \approx 2.18
\]

4. **log₅(100)**:
\[
\log_5(100) = \frac{\log_{10}(100)}{\log_{10}(5)} \approx \frac{2}{0.699} \approx 2.86
\]

5. **log₇(965)**:
\[
\log_7(965) = \frac{\log_{10}(965)}{\log_{10}(7)} \approx \frac{2.985}{0.845} \approx 3.53
\]

6. **log₄(1256)**:
\[
\log_4(1256) = \frac{\log_{10}(1256)}{\log_{10}(4)} \approx \frac{3.100}{0.602} \approx 5.15
\]

7. **log₆(7776)**:
\[
\log_6(7776) = \frac{\log_{10}(7776)}{\log_{10}(6)} \approx \frac{3.891}{0.778} \approx 4.99
\]

### Summary Estimations:

1. log₄(512) ≈ 4.50
2. log₈(200) ≈ 2.55
3. log₆(50) ≈ 2.18
4. log₅(100) ≈ 2.86
5. log₇(965) ≈ 3.53
6. log₄(1256) ≈ 5.15
7. log₆(7776) ≈ 4.99

These are approximate values for the logarithms you provided. If you need more precision, please use a scientific calculator for exact values.