Certainly! Let's rewrite each of the given exponential equations in logarithmic form.
### Part (a)
Given the exponential equation:
[tex]\[ 81^2 = X \][/tex]
We can rewrite this in logarithmic form. The logarithmic form of \( a^b = c \) is:
[tex]\[ \log_a(c) = b \][/tex]
So, applying this to our equation \( 81^2 = X \), we have:
[tex]\[ \log_{81}(X) = 2 \][/tex]
### Part (b)
Given the exponential equation:
[tex]\[ W = 75^{1 / 3} \][/tex]
Again, we rewrite this in logarithmic form using \( a^b = c \) which gives:
[tex]\[ \log_a(c) = b \][/tex]
So, for \( W = 75^{1 / 3} \), we have:
[tex]\[ \log_{75}(W) = \frac{1}{3} \][/tex]
### Part (c)
Given the exponential equation:
[tex]\[ 144 = 12^y \][/tex]
Using the same concept, \( a^b = c \) translates into:
[tex]\[ \log_a(c) = b \][/tex]
So, for \( 144 = 12^y \), we have:
[tex]\[ \log_{12}(144) = y \][/tex]
### Summary
Here are all the exponential equations rewritten in their logarithmic forms:
1. \( 81^2 = X \) rewritten as \( \log_{81}(X) = 2 \)
2. \( W = 75^{1 / 3} \) rewritten as \( \log_{75}(W) = \frac{1}{3} \)
3. \( 144 = 12^y \) rewritten as \( \log_{12}(144) = y \)
This completes the conversion of each given exponential equation to its corresponding logarithmic form.
