Which of the following uses the properties of logarithms to correctly rewrite f(x)=log2(32x^6)

Question

?(1 point)
Responses

f(x)=30 log2 x
f of x is equal to 30 log sub 2 x

f(x)=6 log2 x+5
f of x is equal to 6 log sub 2 x plus 5

f(x)=log2 6x+5
f of x is equal to the log base the log base 2 of 6 of x plus 5

f(x)=6 log2 x+30

Answer 1

To rewrite the function \( f(x) = \log_2(32x^6) \) using the properties of logarithms, we can apply the following properties:

1. Product Property: \( \log_b(mn) = \log_b m + \log_b n \)
2. Power Property: \( \log_b(m^n) = n \log_b m \)

Now, applying these properties to the logarithm:

1. Rewrite \( 32x^6 \) as \( 32 \cdot x^6 \).

   Using the product property: \[ f(x) = \log_2(32) + \log_2(x^6) \]

2. Now apply the power property to \( \log_2(x^6) \): \[ \log_2(x^6) = 6 \log_2(x) \]

3. Next, we need to find \( \log_2(32) \): Since \( 32 = 2^5 \), we have: \[ \log_2(32) = 5 \]

Putting everything together: \[ f(x) = \log_2(32) + \log_2(x^6) = 5 + 6 \log_2(x) \]

Thus, the correct rewriting of \( f(x) \) is: \[ f(x) = 6 \log_2(x) + 5 \]

So the correct option is:

• \( f(x) = 6 \log_2 x + 5 \) (which is the second response).