Answer :
Sure, let's evaluate \(\log_7 98\) given that \(\log_7 2 \approx 0.356\).
To begin with, we will utilize the product property of logarithms, which states that:
[tex]\[ \log_b (xy) = \log_b x + \log_b y \][/tex]
Using this property, we can rewrite \(\log_7 98\) as follows:
[tex]\[ \log_7 98 = \log_7 (2 \cdot 49) \][/tex]
Applying the product property, we get:
[tex]\[ \log_7 98 = \log_7 2 + \log_7 49 \][/tex]
Given: \(\log_7 2 \approx 0.356\),
so we already have:
[tex]\[ \log_7 2 \approx 0.356 \][/tex]
Now, let's evaluate \(\log_7 49\). Notice that \(49\) can be rewritten as \(7^2\):
[tex]\[ \log_7 49 = \log_7 (7^2) \][/tex]
Using the power property of logarithms, which states that \(\log_b (a^c) = c \cdot \log_b a\), we get:
[tex]\[ \log_7 (7^2) = 2 \cdot \log_7 7 \][/tex]
Knowing that \(\log_7 7 = 1\) (since the logarithm of a number to its own base is always 1), we obtain:
[tex]\[ \log_7 49 = 2 \cdot 1 = 2 \][/tex]
Now that we have both \(\log_7 2\) and \(\log_7 49\), we can sum them up:
[tex]\[ \log_7 98 = \log_7 2 + \log_7 49 \approx 0.356 + 2 = 2.356 \][/tex]
So the steps to evaluate \(\log_7 98\) given \(\log_7 2 \approx 0.356\) are:
1. Use the product property: \(\log_7 98 = \log_7 2 + \log_7 49\)
2. Recognize \(49 = 7^2\) and calculate \(\log_7 49 = 2\)
3. Sum the values: \(\log_7 98 \approx 0.356 + 2 = 2.356\)
Therefore, the evaluated logarithm is:
[tex]\[ \log_7 98 \approx 2.356 \][/tex]
Final results:
[tex]\[ \log_7 2 \approx 0.356 \][/tex]
[tex]\[ \log_7 49 = 2 \][/tex]
[tex]\[ \log_7 98 \approx 2.356 \][/tex]
To begin with, we will utilize the product property of logarithms, which states that:
[tex]\[ \log_b (xy) = \log_b x + \log_b y \][/tex]
Using this property, we can rewrite \(\log_7 98\) as follows:
[tex]\[ \log_7 98 = \log_7 (2 \cdot 49) \][/tex]
Applying the product property, we get:
[tex]\[ \log_7 98 = \log_7 2 + \log_7 49 \][/tex]
Given: \(\log_7 2 \approx 0.356\),
so we already have:
[tex]\[ \log_7 2 \approx 0.356 \][/tex]
Now, let's evaluate \(\log_7 49\). Notice that \(49\) can be rewritten as \(7^2\):
[tex]\[ \log_7 49 = \log_7 (7^2) \][/tex]
Using the power property of logarithms, which states that \(\log_b (a^c) = c \cdot \log_b a\), we get:
[tex]\[ \log_7 (7^2) = 2 \cdot \log_7 7 \][/tex]
Knowing that \(\log_7 7 = 1\) (since the logarithm of a number to its own base is always 1), we obtain:
[tex]\[ \log_7 49 = 2 \cdot 1 = 2 \][/tex]
Now that we have both \(\log_7 2\) and \(\log_7 49\), we can sum them up:
[tex]\[ \log_7 98 = \log_7 2 + \log_7 49 \approx 0.356 + 2 = 2.356 \][/tex]
So the steps to evaluate \(\log_7 98\) given \(\log_7 2 \approx 0.356\) are:
1. Use the product property: \(\log_7 98 = \log_7 2 + \log_7 49\)
2. Recognize \(49 = 7^2\) and calculate \(\log_7 49 = 2\)
3. Sum the values: \(\log_7 98 \approx 0.356 + 2 = 2.356\)
Therefore, the evaluated logarithm is:
[tex]\[ \log_7 98 \approx 2.356 \][/tex]
Final results:
[tex]\[ \log_7 2 \approx 0.356 \][/tex]
[tex]\[ \log_7 49 = 2 \][/tex]
[tex]\[ \log_7 98 \approx 2.356 \][/tex]
