Question

a logarithmic function contains the input - output pairs of (4, 4) and (16, 7). the function has the form of ( g(x)=3cdotlog_{b}x + d ) where ( b ) is the base and ( d ) is a constant.
a) write two equations that could be used to find the values of ( b ) and ( d ).
b) find the values of ( b ) and ( d ).

Explanation:

Part a)

Step1: Use point (4, 4)

The function is \( g(x) = 3 \cdot \log_b x + d \). Substitute \( x = 4 \) and \( g(4) = 4 \) into the function.
\( 4 = 3 \cdot \log_b 4 + d \)

Step2: Use point (16, 7)

Substitute \( x = 16 \) and \( g(16) = 7 \) into the function.
\( 7 = 3 \cdot \log_b 16 + d \)

Step1: Subtract the two equations

We have the system:
\(

$$\begin{cases} 4 = 3\log_b 4 + d \\ 7 = 3\log_b 16 + d \end{cases}$$

\)
Subtract the first equation from the second:
\( 7 - 4 = 3\log_b 16 + d - (3\log_b 4 + d) \)
Simplify: \( 3 = 3(\log_b 16 - \log_b 4) \)
Divide both sides by 3: \( 1 = \log_b \frac{16}{4} \) (using the logarithm property \( \log_a M - \log_a N = \log_a \frac{M}{N} \))
So \( 1 = \log_b 4 \)

Step2: Solve for \( b \)

By the definition of logarithms, if \( \log_b 4 = 1 \), then \( b^1 = 4 \), so \( b = 4 \)

Step3: Solve for \( d \)

Substitute \( b = 4 \) and \( x = 4 \), \( g(4) = 4 \) into \( g(x) = 3\log_b x + d \)
\( 4 = 3\log_4 4 + d \)
Since \( \log_4 4 = 1 \), we have \( 4 = 3(1) + d \)
Simplify: \( 4 = 3 + d \)
Subtract 3 from both sides: \( d = 4 - 3 = 1 \)

Answer:

The two equations are \( 4 = 3\log_b 4 + d \) and \( 7 = 3\log_b 16 + d \)

Part b)