Algebra II - Natural Log

Determine the value of: $6ln(e^{-3x})$

$-\frac{1}{2}x$

Explanation

The natural log has a default base of .

According to the rule of logs, we can use:

$log_{b}(b^x) = x$

$6ln(e^{-3x}) = 6ln_e(e^{-3x})$

The coefficient in front of the natural log can be transferred as the power of the exponent.

$6ln_e(e^{-3x}) = ln_e(e^{(-3x)(6)})$

The natural log and base e will cancel, leaving just the exponent.

The answer is:

Evaluate:

$6e^{18}-e^6$

Explanation

Simplify the first term. The natural log has a default base of .

$6e^6 \cdot ln_{e}(e^3)$

According to log rules:

This means that: $6e^6 \cdot ln_{e}(e^3) = 6e^6 \cdot 3 = 18e^6$

The answer is:

Simplify:

Explanation

Use the log properties to separate each term. When the terms inside are multiplied, the logs can be added.

Rewrite the expression.

The exponent, 7 can be dropped as the coefficient in front of the natural log. Natural log of the exponential is equal to one since the natural log has a default base of .

$log_{b}(b^x) = x$

The answer is:

What are the domain and the range of the function ?

Domain = all positive real numbers

Range = all real numbers

Domain = all positive numbers

Range = all non-negative numbers

Domain = all real numbers

Range = all real numbers

Domain = all non-negative numbers

Range = all positive numbers

Domain = all positive numbers

Range = all positive numbers

Explanation

Remember that is still a logarithm of a positive number, .

It's not possible to raise to ANY power and obtain a negative number. Because even $e^{-3}$ , for example, is just $\frac{1}{e^3}$ , which is a ratio of two positive numbers, and therefore positive.

More than that, it's also not possible to obtain 0 by raising to any power. Think: "To what power can I exponentiate e and obtain 0?"

So the domain is strictly positive. It excludes negative numbers and 0.

What about the range? To what possible values are we allowed to exponentiate ?

Well, we just saw that has a definition for negative numbers. (this fact is true for ALL numbers, not just ).

And we can obviously raise it to positive powers. So the range is all real numbers. It includes negative numbers, 0, and positive numbers.

Which of the following expressions is equal to the expression $\ln \left ( x^{2} + 4x + 3\right )$ ?

$\ln (x + 1) + \ln (x + 3)$

$\ln (x + 1) \cdot \ln (x + 3)$

$\ln x^{2} + \ln 4x + \ln 3$

None of the other responses is correct.

$\ln x^{2} \cdot \ln 4x \cdot \ln 3$

Explanation

By the reverse-FOIL method, we factor the polynomial as follows:

$x^{2} + 4x + 3 = (x + 1)(x + 3)$

Therefore, we can use the property

$\ln ab = \ln a + \ln b$

as follows:

$\ln \left ( x^{2} + 4x + 3\right ) =\ln (x + 1)(x + 3) = \ln (x + 1) + \ln (x+3)$

Evaluate: $2e^2ln(e^{2e})$

Explanation

The natural log has a default base of .

Use the log property:

We can cancel the base and the log of the base.

The expression $2e^2ln(e^{2e})$ becomes:

The answer is:

Solve the expression:

$\frac{e^{13}-e}{2}$

$\frac{e^{12}}{2}$

$\frac{e^6}{2}$

$\frac{13-e}{2e}$

$\frac{e^{13}+e}{2}$

Explanation

In order to eliminate the natural log and solve for x, we will need to exponential both sides because is the base of natural log.

$e^{ln(2x+e)}= e^{13}$

The left side will be reduced to just the inner quantity of the natural log.

$2x+e = e^{13}$

Subtract from both sides of the equation.

$2x+e -e= e^{13}-e$

$2x= e^{13}-e$

Divide by two on both sides.

$\frac{2x}{2}= \frac{e^{13}-e}{2}$

The answer is: $\frac{e^{13}-e}{2}$

Solve: $3ln(e^{7e})$

Explanation

The natural log has a default base of . This means that the natural log of to the certain power will be just the power itself.

$log_{b}(b^x) = x$

The expression $3ln(e^{7e})$ becomes:

The answer is:

Solve for x:

$\ln e^{x}+\ln e^{5}=1$

$x=e^{-4}$

$x=e^{4}$

Explanation

To solve for x, keep in mind that the natural logarithm and the exponential cancel each other out (property of any logarithm with a base that is being taken of that same base with an exponent attached). When they cancel, we are just left with the exponents: