Question

question 11 of 25
find f(5) for f(x) = \frac{1}{9}(3)^x.

a. 9
b. 81
c. 27
d. 3

Explanation:

Step1: Substitute x=5 into the function

We have the function \( f(x)=\frac{1}{9}(3)^{x} \). To find \( f(5) \), we substitute \( x = 5 \) into the function. So we get \( f(5)=\frac{1}{9}(3)^{5} \).

Step2: Calculate \( 3^{5} \)

First, calculate \( 3^{5} \). We know that \( 3^{5}=3\times3\times3\times3\times3 = 243 \).

Step3: Multiply by \( \frac{1}{9} \)

Now, multiply \( \frac{1}{9} \) by \( 243 \). So \( \frac{1}{9}\times243=\frac{243}{9} = 27 \). Wait, no, wait, \( 3^5 = 243 \), \( \frac{1}{9}\times243 = 27 \)? Wait, no, wait, \( 3^2 = 9 \), so \( 3^5=3^{2 + 3}=3^2\times3^3 = 9\times27 = 243 \). Then \( \frac{1}{9}\times3^5=\frac{3^5}{3^2}=3^{5 - 2}=3^3 = 27 \)? Wait, no, wait, the original function is \( \frac{1}{9}(3)^x=\frac{3^x}{3^2}=3^{x - 2} \). So when \( x = 5 \), \( 3^{5 - 2}=3^3 = 27 \)? Wait, but let's recalculate. Wait, \( \frac{1}{9}\times3^5=\frac{3^5}{9}=\frac{243}{9}=27 \). But wait, the options have 27 as option C? Wait, no, the options are A.9, B.81, C.27, D.3. Wait, but let's check again. Wait, \( 3^5 = 243 \), \( 243\div9 = 27 \). So \( f(5)=27 \), which is option C. Wait, but wait, maybe I made a mistake. Wait, \( \frac{1}{9}(3)^5=\frac{3^5}{3^2}=3^{3}=27 \). Yes, that's correct.

Answer:

C. 27