Question

I need help with this problem to help my son

Answer 1

`\begin{gathered} log_(1.1)((400)/(100))=x \\ \end{gathered}`

The frog enclosure will reach maximum capacity after approximately 15 weeks.

Explanations:

The standard exponential equation is in the form;

`f(x)=ab^x`

Using the coordinate points (0, 100) and (1, 110) from the table, we can create a system of equation as shown:

`\begin{gathered} 100=ab^0 \\ 110=ab^1 \end{gathered}`

Divide both equations to have:

`\begin{gathered} (110)/(100)=(ab^1)/(ab^0) \\ 1.1=b^(1-0) \\ b=1.1 \end{gathered}`

Determine the value of a by substitute the value of b into any of the equation as shown:

`\begin{gathered} 110=ab^1 \\ ab=110 \\ 1.1a=110 \\ a=100 \end{gathered}`

Determine the equivalent exponential function

`\begin{gathered} f(x)=ab^x \\ f(x)=100(1.1)^x \end{gathered}`

If the maximum capacity is 400 frogs, then f(x) = 400. On substituting into the resulting function, we will have:

`\begin{gathered} 400=100(1.1)^x \\ 1.1^x=(400)/(100) \end{gathered}`

Converting the expression into logarithm function

`\begin{gathered} log_(1.1)((400)/(100))=x \\ x=log_(1.1)((400)/(100)) \end{gathered}`

Determine the value of x

`\begin{gathered} 1.1^x=(400)/(100) \\ 1.1^x=4 \\ xlog1.1=log4 \\ x=(log4)/(log1.1) \\ x=(1.6021)/(1.0414) \\ x\approx15weeks \end{gathered}`

Hence the frog enclosure will reach maximum capacity after approximately 15 weeks.