Which student correctly rewrote the function f(x)=log11(x−9)

in exponential form?

Student #1: y11+9=x


Student #2: 11y+9=x


Student #3: 11y+9=x


(1 point)

1 answer

To rewrite the function \( f(x) = \log_{11}(x - 9) \) in exponential form, we start by recalling the relationship between logarithms and exponents. The logarithmic equation \( \log_{b}(a) = c \) is equivalent to the exponential form \( b^c = a \).

In this case, we can set the function equal to \( y \): \[ y = \log_{11}(x - 9) \]

This implies: \[ 11^y = x - 9 \]

To rearrange this into a form that isolates \( x \), we add 9 to both sides: \[ x = 11^y + 9 \]

Now we can analyze the students' responses:

  • Student #1: \( y11 + 9 = x \) - This is not correctly written since \( y \cdot 11 + 9 \) does not match the form we derived.

  • Student #2: \( 11y + 9 = x \) - This is also not correct; it seems to imply multiplication rather than an exponent.

  • Student #3: \( 11^y + 9 = x \) - This is the correct form we derived.

Thus, Student #3 correctly rewrote the function \( f(x) = \log_{11}(x - 9) \) in exponential form.