Question

1. selected values of the logarithmic function g are given in the table above, where ( g(x) = a + b ln x ).

a) use the data to write two equations that can be used to find the values for constants a and b in the expression for ( g(x) ).
b) find the values of a and b.

Explanation:

Part (a)

Step1: Substitute \( x = 3 \), \( g(x)=11 \)

We know \( g(x)=a + b\ln x \). Substituting \( x = 3 \) and \( g(3)=11 \) into the function, we get the equation:
\( 11=a + b\ln 3 \)

Step2: Substitute \( x = 10 \), \( g(x)=14.5 \)

Substituting \( x = 10 \) and \( g(10)=14.5 \) into \( g(x)=a + b\ln x \), we get the second equation:
\( 14.5=a + b\ln 10 \)

Part (b)

Step1: Subtract the two equations

We have the system of equations:
\(

$$\begin{cases} 11=a + b\ln 3 \quad (1)\\ 14.5=a + b\ln 10 \quad (2) \end{cases}$$

\)

Subtract equation (1) from equation (2):
\( 14.5 - 11=(a + b\ln 10)-(a + b\ln 3) \)
Simplify the right - hand side: \( a - a + b(\ln 10-\ln 3)=b\ln\frac{10}{3} \) (using the property \( \ln m-\ln n=\ln\frac{m}{n} \))
The left - hand side is \( 3.5 \), so \( 3.5 = b\ln\frac{10}{3} \)

Solve for \( b \):
\( b=\frac{3.5}{\ln\frac{10}{3}} \approx\frac{3.5}{\ln\frac{10}{3}}\approx\frac{3.5}{1.20397}\approx2.907 \) (we can also keep it as \( b = \frac{7/2}{\ln(10/3)}=\frac{7}{2\ln(10/3)} \))

Step2: Substitute \( b \) back to find \( a \)

Substitute \( b \) into equation (1): \( 11=a + b\ln 3 \)
\( a=11 - b\ln 3 \)

Substitute \( b=\frac{3.5}{\ln\frac{10}{3}} \) into the above formula:
\( a = 11-\frac{3.5\ln 3}{\ln\frac{10}{3}} \)

We know that \( \ln 3\approx1.0986 \), \( \ln\frac{10}{3}\approx1.20397 \)

\( 3.5\ln 3\approx3.5\times1.0986 = 3.8451 \)

\( \frac{3.8451}{1.20397}\approx3.194 \)

\( a=11 - 3.194 = 7.806 \)

Part (a) Answer: The two equations are \( \boldsymbol{11=a + b\ln 3} \) and \( \boldsymbol{14.5=a + b\ln 10} \)

Part (b) Answer: \( a\approx\boldsymbol{7.81} \), \( b\approx\boldsymbol{2.91} \) (or more precise values depending on the method of calculation)

Answer:

Step1: Subtract the two equations

We have the system of equations:
\(

$$\begin{cases} 11=a + b\ln 3 \quad (1)\\ 14.5=a + b\ln 10 \quad (2) \end{cases}$$

\)

Subtract equation (1) from equation (2):
\( 14.5 - 11=(a + b\ln 10)-(a + b\ln 3) \)
Simplify the right - hand side: \( a - a + b(\ln 10-\ln 3)=b\ln\frac{10}{3} \) (using the property \( \ln m-\ln n=\ln\frac{m}{n} \))
The left - hand side is \( 3.5 \), so \( 3.5 = b\ln\frac{10}{3} \)

Solve for \( b \):
\( b=\frac{3.5}{\ln\frac{10}{3}} \approx\frac{3.5}{\ln\frac{10}{3}}\approx\frac{3.5}{1.20397}\approx2.907 \) (we can also keep it as \( b = \frac{7/2}{\ln(10/3)}=\frac{7}{2\ln(10/3)} \))

Step2: Substitute \( b \) back to find \( a \)

Substitute \( b \) into equation (1): \( 11=a + b\ln 3 \)
\( a=11 - b\ln 3 \)

Substitute \( b=\frac{3.5}{\ln\frac{10}{3}} \) into the above formula:
\( a = 11-\frac{3.5\ln 3}{\ln\frac{10}{3}} \)

We know that \( \ln 3\approx1.0986 \), \( \ln\frac{10}{3}\approx1.20397 \)

\( 3.5\ln 3\approx3.5\times1.0986 = 3.8451 \)

\( \frac{3.8451}{1.20397}\approx3.194 \)

\( a=11 - 3.194 = 7.806 \)

Part (a) Answer: The two equations are \( \boldsymbol{11=a + b\ln 3} \) and \( \boldsymbol{14.5=a + b\ln 10} \)

Part (b) Answer: \( a\approx\boldsymbol{7.81} \), \( b\approx\boldsymbol{2.91} \) (or more precise values depending on the method of calculation)