Question

regression equation: $y = 3.915(1.106)^x$
the pond can hold 400 water lilies. by what day will the pond be full? write and solve an equation.

1. the pond will be full by the end of day \boxed{46}

complete
which solution method did you use?
logarithms
graphing $y = 3.915(1.106)^x$ and tracing
graphing $y = 3.915(1.106)^x$ and $y = 400$, then finding the x - value of the intersection
complete
by the 46th day, there are 400 water lilies in the pond. the estimate you made
close to 46 (or exactly right) because it was estimated well.
too low because the quick exponential growth was not accounted for.
too high because the exponential growth was overestimated.
done

Explanation:

Step1: Set up the equation

We know the regression equation is \( y = 3.915(1.106)^x \), and the pond can hold 400 water lilies. So we set \( y = 400 \), getting the equation \( 400=3.915(1.106)^x \).

Step2: Solve for \( x \)

First, divide both sides by 3.915: \( \frac{400}{3.915}=(1.106)^x \). Calculate \( \frac{400}{3.915}\approx102.17 \). So we have \( 102.17=(1.106)^x \).
Then, take the natural logarithm of both sides: \( \ln(102.17)=x\ln(1.106) \).
We know that \( \ln(102.17)\approx4.627 \) and \( \ln(1.106)\approx0.101 \).
Now, solve for \( x \): \( x = \frac{\ln(102.17)}{\ln(1.106)}=\frac{4.627}{0.101}\approx45.81 \approx 46 \).

Step3: Analyze the solution method

The method used here is using logarithms (specifically natural logarithms) to solve the exponential equation. We transformed the exponential equation into a linear - like form by taking logarithms, which is a standard method for solving exponential equations of the form \( a = b^x \) (or in this case, \( a = c\cdot b^x \)).

Answer:

The solution method used is Logarithms. To solve \( 400 = 3.915(1.106)^x \), we first divide by 3.915 to get \( \frac{400}{3.915}=(1.106)^x \), then take the natural logarithm of both sides (\( \ln(102.17)=x\ln(1.106) \)) and solve for \( x \), resulting in \( x\approx46 \).