Linear Algebra Examples

$N = \frac{300}{5 + 24 e^{- 0.2 t}}$

Step 1

Set the denominator in $\frac{300}{5 + 24 e^{- 0.2 t}}$ equal to $0$ to find where the expression is undefined.

$5 + 24 e^{- 0.2 t} = 0$

Step 2

Solve for $t$ .

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Step 2.1

Subtract $5$ from both sides of the equation.

$24 e^{- 0.2 t} = - 5$

Step 2.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (24 e^{- 0.2 t}) = \ln (- 5)$

Step 2.3

The equation cannot be solved because $\ln (- 5)$ is undefined.

Undefined

Step 2.4

There is no solution for $24 e^{- 0.2 t} = - 5$

No solution

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${t | t \in ℝ}$

Step 4