Given the regression equation: [tex][tex]$y=3.915(1.106)^x$[/tex][/tex], the pond can hold 400 water lilies.

By what day will the pond be full? Write and solve an equation.

The pond will be full by the end of day [tex][tex]$\square$[/tex][/tex]

Answer :

To determine by what day the pond will be full of water lilies, we need to solve the regression equation for [tex]\( x \)[/tex] (representing days) when the number of water lilies reaches the pond capacity of 400. The given regression equation is:

[tex]\[ y = 3.915(1.106)^x \][/tex]

We seek the value of [tex]\( x \)[/tex] when [tex]\( y \)[/tex] equals 400. Therefore, we set up the equation:

[tex]\[ 400 = 3.915(1.106)^x \][/tex]

Follow these steps to solve for [tex]\( x \)[/tex]:

1. Isolate the exponential term:
[tex]\[ (1.106)^x = \frac{400}{3.915} \][/tex]

2. Calculate the ratio:
[tex]\[ \frac{400}{3.915} = 102.17113665389527 \][/tex]

3. Take the natural logarithm of both sides to bring the exponent [tex]\( x \)[/tex] down:
[tex]\[ \ln((1.106)^x) = \ln(102.17113665389527) \][/tex]

Using the properties of logarithms, we can simplify the left-hand side:
[tex]\[ x \ln(1.106) = \ln(102.17113665389527) \][/tex]

4. Calculate the logarithms:
[tex]\[ \ln(102.17113665389527) = 4.626649217665215 \][/tex]
[tex]\[ \ln(1.106) = 0.10074990310014315 \][/tex]

5. Solve for [tex]\( x \)[/tex] by dividing the logarithms:
[tex]\[ x = \frac{4.626649217665215}{0.10074990310014315} = 45.92212076935131 \][/tex]

Since we are interested in the day by which the pond will be full, and since the number of lilies can only be counted in whole days, we round up to the nearest whole number:

The pond will be full by the end of day [tex]\( \boxed{46} \)[/tex]

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