Answer:
[tex]\begin{gathered} log_{1.1}(\frac{400}{100})=x \\ \end{gathered}[/tex]
The frog enclosure will reach maximum capacity after approximately 15 weeks.
Explanations:
The standard exponential equation is in the form;
[tex]f(x)=ab^x[/tex]
Using the coordinate points (0, 100) and (1, 110) from the table, we can create a system of equation as shown:
[tex]\begin{gathered} 100=ab^0 \\ 110=ab^1 \end{gathered}[/tex]
Divide both equations to have:
[tex]\begin{gathered} \frac{110}{100}=\frac{ab^1}{ab^0} \\ 1.1=b^{1-0} \\ b=1.1 \end{gathered}[/tex]
Determine the value of a by substitute the value of b into any of the equation as shown:
[tex]\begin{gathered} 110=ab^1 \\ ab=110 \\ 1.1a=110 \\ a=100 \end{gathered}[/tex]
Determine the equivalent exponential function
[tex]\begin{gathered} f(x)=ab^x \\ f(x)=100(1.1)^x \end{gathered}[/tex]
If the maximum capacity is 400 frogs, then f(x) = 400. On substituting into the resulting function, we will have:
[tex]\begin{gathered} 400=100(1.1)^x \\ 1.1^x=\frac{400}{100} \end{gathered}[/tex]
Converting the expression into logarithm function
[tex]\begin{gathered} log_{1.1}(\frac{400}{100})=x \\ x=log_{1.1}(\frac{400}{100}) \end{gathered}[/tex]
Determine the value of x
[tex]\begin{gathered} 1.1^x=\frac{400}{100} \\ 1.1^x=4 \\ xlog1.1=log4 \\ x=\frac{log4}{log1.1} \\ x=\frac{1.6021}{1.0414} \\ x\approx15weeks \end{gathered}[/tex]
Hence the frog enclosure will reach maximum capacity after approximately 15 weeks.
