Math - Solving and Graphing Exponential Equations

The population of a certain bacteria increases exponentially according to the following equation:

$P(t)=2000e^{^{2t}}$

where P represents the total population and t represents time in minutes.

How many minutes does it take for the bacteria's population to reach 48,000?

$\frac{\ln24}{2}$

$\frac{\log 24}{2}$

$\frac{2}{\ln 24}$

$\frac{2}{\log 24}$

$\ln 12$

Explanation

The question gives us P (48,000) and asks us to find t (time). We can substitute for P and start to solve for t:

$48,000 = 2000e^{2t}$

$24 = e^{2t}$

Now we have to isolate t by taking the natural log of both sides:

$\ln 24 = \ln e^{2t}$

$\ln 24 = (2t)\ln e$

And since $\ln e = 1$ , t can easily be isolated:

$\ln 24 = 2t$

$\frac{\ln24}{2} = t$

Note: $\frac{\ln 24}{2}$ does not equal $\ln 12$ . You have to perform the log operation first before dividing.

Find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

and

$x=\frac{1}{8}$

$x=2\sqrt{2}$

$x=2\sqrt{11}$

Explanation

To find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ , we need to set the numerator equal to zero and solve.

First, notice that can be factored into . Now set that equal to zero: .

Since we have two sets in parentheses, there are two separate values that can cause our equation to equal zero: one where and one where .

Solve for each value:

and

Therefore there are two -interecpts: and .

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

$y=\frac{3}{2}$

There are no horizontal asymptotes.

$y=\frac{2}{3}$

Explanation

Since the exponents of the variables in both the numerator and denominator are equal, the horizontal asymptote will be the coefficient of the numerator's variable divided by the coefficient of the denominator's variable.

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

$y=\frac{3}{2}$

There are no horizontal asymptotes.

$y=\frac{2}{3}$

Explanation

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

What are the -intercepts of the equation?

$y=\frac{x^2-9}{x^2-16}$

There are no -intercepts.

Explanation

To find the x-intercepts of the equation, we set the numerator equal to zero.

$\sqrt{9}=\sqrt{x^2}$

$y=\frac{x^2-9}{x^2-16}$

What is the horizontal asymptote of this equation?

There is no horizontal asymptote.

Explanation

Look at the exponents of the variables. Both our numerator and denominator are . Therefore the horizontal asymptote is calculated by dividing the coefficient of the numerator by the coefficient of the denomenator.

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

What are the -intercepts of the equation?

$y=\frac{x^2-9}{x^2-16}$

There are no -intercepts.

Explanation

To find the x-intercepts of the equation, we set the numerator equal to zero.

$\sqrt{9}=\sqrt{x^2}$

Find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ .

and

$x=\frac{1}{8}$

$x=2\sqrt{2}$

$x=2\sqrt{11}$

Explanation

To find the -intercept(s) of $y=\frac{x^2+9x+8}{x^2-44}$ , we need to set the numerator equal to zero and solve.

First, notice that can be factored into . Now set that equal to zero: .

Since we have two sets in parentheses, there are two separate values that can cause our equation to equal zero: one where and one where .

Solve for each value:

and

Therefore there are two -interecpts: and .

$y=\frac{x^2-9}{x^2-16}$

What is the horizontal asymptote of this equation?

There is no horizontal asymptote.

Explanation

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

The population of a certain bacteria increases exponentially according to the following equation:

$P(t)=2000e^{^{2t}}$

where P represents the total population and t represents time in minutes.

How many minutes does it take for the bacteria's population to reach 48,000?

$\frac{\ln24}{2}$

$\frac{\log 24}{2}$

$\frac{2}{\ln 24}$

$\frac{2}{\log 24}$

$\ln 12$

Explanation

The question gives us P (48,000) and asks us to find t (time). We can substitute for P and start to solve for t:

$48,000 = 2000e^{2t}$

$24 = e^{2t}$

Now we have to isolate t by taking the natural log of both sides:

$\ln 24 = \ln e^{2t}$

$\ln 24 = (2t)\ln e$

And since $\ln e = 1$ , t can easily be isolated:

$\ln 24 = 2t$

$\frac{\ln24}{2} = t$

Note: $\frac{\ln 24}{2}$ does not equal $\ln 12$ . You have to perform the log operation first before dividing.

Solving and Graphing Exponential Equations

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