Pre-Calculus - Exponential Functions

Solve: $6\mathrm{e}^{4x}-16=8$

$x=\frac{\ln4}{4}$

$x=\ln4$

$x=\ln\frac{1}{4}$

None of the other answers.

Explanation

$6\mathrm{e}^{4x}-16=8$

Combine the constants:

$6\mathrm{e}^{4x}=24$

Isolate the exponential function by dividing:

$\mathrm{e}^{4x}=4$

Take the natural log of both sides:

$\ln(\mathrm{e}^{4x})=\ln(4)$

$4x=\ln(4)$

Finally isolate x:

$x=\frac{\ln(4)}{4}\hspace{1mm}or\hspace{1mm}.35$

Solve for .

$2e^5e^{-3}\ln(e^x)=2x^2$

$x=\ln(2e)$

$x=2\ln(e^2)$

Explanation

First, let's begin by simplifying the left hand side.

$e^5e^{-3}$ becomes and $\ln(e^x)$ becomes . Remember that $\ln(e)=1$ , and the in that expression can come out to the front, as in $x\cdot \ln(e)$ .

Now, our expression is

From this, we can cancel out the 2's and an x from both sides.

Thus our answer becomes:

Solve: $6\mathrm{e}^{4x}-16=8$

$x=\frac{\ln4}{4}$

$x=\ln4$

$x=\ln\frac{1}{4}$

None of the other answers.

Explanation

$6\mathrm{e}^{4x}-16=8$

Combine the constants:

$6\mathrm{e}^{4x}=24$

Isolate the exponential function by dividing:

$\mathrm{e}^{4x}=4$

Take the natural log of both sides:

$\ln(\mathrm{e}^{4x})=\ln(4)$

$4x=\ln(4)$

Finally isolate x:

$x=\frac{\ln(4)}{4}\hspace{1mm}or\hspace{1mm}.35$

Solve for .

$2e^5e^{-3}\ln(e^x)=2x^2$

$x=\ln(2e)$

$x=2\ln(e^2)$

Explanation

First, let's begin by simplifying the left hand side.

$e^5e^{-3}$ becomes and $\ln(e^x)$ becomes . Remember that $\ln(e)=1$ , and the in that expression can come out to the front, as in $x\cdot \ln(e)$ .

Now, our expression is

From this, we can cancel out the 2's and an x from both sides.

Thus our answer becomes:

Solve the equation for .

$2e^{2x}-7e^{x}+6=0$

$x=ln2,x=ln(\frac{3}{2})$

$x=-ln(\frac{3}{4}),x=ln3$

$x=\pm ln3$

Explanation

$2e^{2x}-7e^{x}+6=0$

The key to this is that $e^{2x}=e^{x}\cdot e^{x}$ . From here, the equation can be factored as if it were $2x^{2}-7x+6=0$ .

$(2e^{x}-3)(e^{x}-2)=0$

$(2e^{x}-3)=0$ and $(e^{x}-2)=0$

$2e^{x}=3$ and $e^{x}=2$

$e^{x}=\frac{3}{2}$ and $e^{x}=2$

Now take the natural log (ln) of the two equations.

$ln(e^{x})=ln(\frac{3}{2})$ and $ln(e^{x})=ln2$

$x=ln\frac{3}{2}$ and

Solve the equation for .

$2e^{2x}-7e^{x}+6=0$

$x=ln2,x=ln(\frac{3}{2})$

$x=-ln(\frac{3}{4}),x=ln3$

$x=\pm ln3$

Explanation

$2e^{2x}-7e^{x}+6=0$

The key to this is that $e^{2x}=e^{x}\cdot e^{x}$ . From here, the equation can be factored as if it were $2x^{2}-7x+6=0$ .

$(2e^{x}-3)(e^{x}-2)=0$

$(2e^{x}-3)=0$ and $(e^{x}-2)=0$

$2e^{x}=3$ and $e^{x}=2$

$e^{x}=\frac{3}{2}$ and $e^{x}=2$

Now take the natural log (ln) of the two equations.

$ln(e^{x})=ln(\frac{3}{2})$ and $ln(e^{x})=ln2$

$x=ln\frac{3}{2}$ and

The population of fish in a pond is modeled by the exponential function

$y=1046e^{-0.1t}$ , where is the population of fish and is the number of years since January 2010.

Determine the population of fish in January 2010 and January 2015.

2010: fish

2015: fish

2010: fish

2015: fish

2010: fish

2015: fish

2010: fish

2015: fish

Explanation

In 2010, in our equation because we have had no years past 2010. Plugging that in to the model equation and solving:

$y=1046e^{-0.1(0)} = 1046$ , since anything raised to the power of zero becomes . So the population of fish in 2010 is fish.

In 2015, because 5 years have passed since 2010. Plugging that into our equation and solving gives us

$y=1046e^{-0.1(5)} \approx 634$

So the population of fish in 2015 is fish. This is an example of exponential decay since the function is decreasing.

The population of fish in a pond is modeled by the exponential function

$y=1046e^{-0.1t}$ , where is the population of fish and is the number of years since January 2010.

Determine the population of fish in January 2010 and January 2015.

2010: fish

2015: fish

2010: fish

2015: fish

2010: fish

2015: fish

2010: fish

2015: fish

Explanation

In 2010, in our equation because we have had no years past 2010. Plugging that in to the model equation and solving:

$y=1046e^{-0.1(0)} = 1046$ , since anything raised to the power of zero becomes . So the population of fish in 2010 is fish.

In 2015, because 5 years have passed since 2010. Plugging that into our equation and solving gives us

$y=1046e^{-0.1(5)} \approx 634$

So the population of fish in 2015 is fish. This is an example of exponential decay since the function is decreasing.

Choose the description that matches the equation below:

$y = (\frac{1}{3})^{4x}$

Exponential decay

$y = \frac{1}{3}$

Exponential decay

Exponential growth

Explanation

Exponential graphs can either decay or grow. This is based on the value of the base of the exponent. If the base is greater than , the graph will be growth. And, if the base is less than , then the graph will be decay. In this situation, our base is $\frac{1}{3}$ . Since this is less than , we have a decay graph. Then, to determine the y-intercept we substitute . Thus, we get: