Natural Log - Algebra II

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Question

$5^{x}=2e^5$

Solve for .

Answer

The first thing we notice about this problem is that is an exponent. This should be an immediate reminder: use logs!

The question is, which base should we choose for the log? We should use the natural log (log base e) because the right-hand side of the equation already has e as a base of an exponent. As you will see, things cancel out more nicely this way.

$5^{x}=2e^5$

Take the natural log of both sides:

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

Rewrite the right-hand side of the equation using the product rule for logs:

$\textup{ln}5^x = \textup{ln}2+\textup{ln}e^5$

Now rewrite the whole equation after bringing down those exponents.

$x\textup{ln}5 = \textup{ln}2+5\textup{ln}e$

$\textup{ln}e$ is the same thing as $\textup{log}_{e}e$ , which equals 1.

$x\textup{ln}5 = \textup{ln}2+5$

Now we just divide by $\textup{ln}5$ on both sides to isolate .

$x = \frac{\textup{ln}2+5}{\textup{ln}5}$

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Question

Rewrite as a single logarithmic expression:

$2 \ln x - \ln (x + 1)$

Answer

Using the properties of logarithms

$n \ln a = \ln a^{n}$ and $\ln a - \ln b = \ln \frac{a}{b}$ ,

we simplify as follows:

$2 \ln x - \ln (x + 1)$

$= \ln x^{2} - \ln (x + 1)$

$= \ln \frac{x^{2} }{x + 1}$

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Question

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

Answer

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:

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

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Question

What are the domain and the range of the function ?

Answer

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.

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Question

Solve $6e^{-x}+1=3$ . Round to the nearest thousandth.

Answer

The original equation is:

$6e^{-x}+1=3$

Subtract from both sides:

$6e^{-x}=2$

Divde both sides by :

$e^{-x}=\frac{2}{6}=\frac{1}{3}$

Take the natural logarithm of both sides:

$-x=\ln \frac{1}{3}$

Divde both sides by and use a calculator to get:

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Question

Solve for :

.

If necessary, round to the nearest tenth.

Answer

Give both sides the same base, using e:

$e^{ln(x+3)}=e^3$ .

Because e and ln cancel each other out, .

Solve for x and round to the nearest tenth:

$x\approx17.1$

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Question

Solve for x:

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

Answer

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:

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Question

Determine the value of: $3ln(e^{x^2-1})$

Answer

The natural log has a base of . This means that the term will simplify to whatever is the power of . Some examples are:

This means that $ln(e^{x^2-1}) = (x^2-1)$

Multiply this quantity with three.

The answer is:

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Question

Determine the value of: $2e\cdot ln(e^3)$

Answer

In order to simplify this expression, use the following natural log rule.

The natural log has a default base of . This means that:

$2e\cdot ln(e^3)=2e\cdot ln_e(e^3) = 2e\cdot (3)$

The answer is:

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Question

Simplify: $2e^{4ln{(2e)}}$

Answer

According to log properties, the coefficient in front of the natural log can be rewritten as the exponent raised by the quantity inside the log.

$2e^{4ln{(2e)}} = 2e^{ln[{(2e)^4}]}$

Notice that natural log has a base of . This means that raising the log by base will eliminate both the and the natural log.

The terms become:

Simplify the power.

The answer is:

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Question

Evaluate:

Answer

The natural log has a default base of . Natural log to of an exponential raised to the power will be just the power. The natural log and will be eliminated.

Rewrite the expression.

The answer is:

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Question

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

Answer

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:

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Question

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

Answer

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:

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Question

Simplify:

Answer

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:

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Question

The equation $kt = ln(\frac{T-S}{T_0-S})$ represents Newton's Law of Cooling. Solve for .

Answer

Use base and raise both the left and right sides as the powers of . This will eliminate the natural log term.

$e^{kt} =e^ {ln(\frac{T-S}{T_0-S})}$

The equation becomes:

$e^{kt} =\frac{T-S}{T_0-S}$

Multiply the quantity on both sides.

$e^{kt}(T_0-S) =\frac{T-S}{T_0-S}(T_0-S)$

$e^{kt}T_0-e^{kt}S =T-S$

Add $e^{kt}S$ on both sides.

$e^{kt}T_0-e^{kt}S +e^{kt}S=T-S+e^{kt}S$

The equation becomes:

$e^{kt}T_0=T-S+e^{kt}S$

To isolate , divide $e^{kt}$ on both sides.

$\frac{e^{kt}T_0}{e^{kt}}=\frac{T-S+e^{kt}S}{e^{kt}}$

The answer is: $T_0 =\frac{T-S+e^{kt}S}{e^{kt}}$

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Question

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

Answer

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:

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Question

Simplify:

Answer

Notice that the terms inside the log are added, with a common factor of . Pull out a common factor.

Notice that these two terms inside the log are multiplied. We can split the log into two terms.

The value of the first term is equal to one, since natural log has a default base of . We can use the property to eliminate the log and the term, which will cancel leaving just the power of one.

The expression becomes:

We cannot use the property of logs to simplify the second term.

The answer is:

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Question

Solve the expression:

Answer

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

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Question

Evaluate:

Answer

The natural log has a default base of .

This means that: $ln_{e}(e^2)$

According to the property of logs, $log_{a}a^x = x$ , and will equal just the power.

Simplify the expression.

The answer is:

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Question

Evaluate:

Answer

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:

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