How to Cancel Out a Natural Log (ln)

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Solving a natural log is a breeze with this easy, understandable guide

Co-authored by David Jia and Devin McSween

Last Updated: September 9, 2024 Fact Checked

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What is ln?

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Canceling Out ln

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Properties of ln

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Getting Rid of e

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Video

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Q&A

Solving an equation for x when a natural logarithm is involved can definitely leave you scratching your head. That little ln is stopping you from getting x alone, so how do you get rid of it? Luckily, solving a natural logarithm is easier than it looks! In this article, we’ll tell you what a natural log is and show you how to get rid of it in an equation. If you’re ready to solve some math problems, read on!

Things You Should Know

The natural log, denoted as $ln$ , is a logarithm with the base $e.$
$Ln(x)$ and $e^{x}$ are inverse functions, so raise $ln(x)$ to the base of $e$ to get rid of $ln.$
To solve an equation with $ln(x)$ , isolate $ln(x)$ to 1 side. Combine any natural logs together. Then, exponentiate each side with $e$ to cancel out $ln$ .

Section 1 of 4:

What is ln?

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Natural log, written as $ln$ , is a logarithm that uses the base of $e$ . A logarithm is the inverse of an exponent. It tells you what a base number has to be raised to in order to produce another number. Natural logs, written as $ln$ $ln$ always use the base number $e$ $e$ , which is an irrational number that is approximately equal to 2.718.^{[1]
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- In simpler terms, logs and natural logs ask what exponent produced the answer.
- Because logs and exponents are inverses of each other, writing $log2(8)=3$ is the same thing as writing $2^{3}=8$ . The base of the logarithm is 2, which is raised to the power of 3 to produce 8.
- Using a natural log example, $ln(30)=3.4012$ is another way of writing $e^{3.4012}=30$ . The base $e$ is raised to 3.4012 to produce 30.

Section 2 of 4:

Canceling Out ln

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1
Isolate the natural log on 1 side of the equation. If your equation has multiple variables, move them around so the natural log is by itself on one side. In equations with multiple natural logs, move them to the same side of the equation and combine them.^{[2]
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- For example, solve the equation $3+2ln({\frac {x}{7}}+3)=-4.$ $3+2ln({\frac {x}{7}}+3)=-4.$
  - Subtract the 3 to move it to the other side: $2ln({\frac {x}{7}}+3)=-7.$
  - Divide by 2 to fully isolate ln: $ln({\frac {x}{7}}+3)={\frac {-7}{2}}.$
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Exponentiate both sides of the equation with $e$ to cancel out $ln$ . To get rid of the natural log in an expression, simply raise each side of the equation to the base of $e$ $e$ . $Ln$ $Ln$ and $e$ $e$ cancel out, leaving you with a solvable expression.^{[3]
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- Using the simplified equation $ln({\frac {x}{7}}+3)={\frac {-7}{2}}$ $ln({\frac {x}{7}}+3)={\frac {-7}{2}}$ :
  - Exponentiate each side with $e$ : $e^{ln({\frac {x}{7}}+3)}=e^{\frac {-7}{2}}.$
  - Cancel out $ln$ : ${\frac {x}{7}}+3=e^{\frac {-7}{2}}.$
3
Solve the equation for x. With $ln$ $ln$ gone from the equation, solve it like normal. An $e$ $e$ is often still in the equation, so follow your teacher’s directions to either leave it in the final solution, or simplify the answer in decimals. To get the answer in decimals, use a scientific calculator.^{[4]
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- Solve ${\frac {x}{7}}+3=e^{\frac {-7}{2}}$ ${\frac {x}{7}}+3=e^{\frac {-7}{2}}$ :
  - Subtract 3 from both sides: ${\frac {x}{7}}=e^{\frac {-7}{2}}-3.$
  - Multiply 7 by both sides: $x=7(e^{\frac {-7}{2}}-3).$ Leave the equation like this, or simplify.
  - Round the answer to the nearest thousandth using a calculator: $x=7(e^{\frac {-7}{2}}-3)=-20.789.$
$Step 4 Practice getting rid of l n {\displaystyle ln} with multiple logs.$

4
Practice getting rid of $ln$ with multiple logs. In more complicated expressions, your equation might have more than 1 natural log. Just follow the same steps as above, making sure to combine the natural logs before you exponentiate.^{[5]
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- Solve the equation $ln(x-1)=1+ln(3x+2).$ $ln(x-1)=1+ln(3x+2).$
  - Isolate $ln$ to 1 side of the equation: $ln(x-1)-ln(3x+2)=1.$
  - Combine the natural logs together. In this equation, 2 logs are subtracted. So, combine them by dividing them by each other. $ln({\frac {x-1}{3x+2}})=1.$
  - Exponentiate both sides with $e$ : $e^{ln({\frac {x-1}{3x+2}})}=e^{1}.$
  - Cancel out $ln$ : ${\frac {x-1}{3x+2}}=e.$
  - Simplify and solve:
    $x-1=e(3x+2)$
    
    $x-1=3ex+2e$
    
    $x-1-3ex=2e$
    
    $x-3ex=2e+1$
    
    $(1-3e)x=2e+1$
    
    $x={\frac {2e+1}{1-3e}}$
  - Round the answer to the nearest thousandth: $x=-0.900.$

Section 3 of 4:

Properties of ln and Logs

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1
Logs and exponents are inverse functions that undo each other. Because natural logs and $e$ $e$ are closely related, all you have to do is raise $ln$ $ln$ to the base of $e$ $e$ to get rid of it. Similarly, if you have an equation with $e$ $e$ , you take its natural log to cancel it out.^{[6]
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- For example, if you have $ln(x),$ you raise it to the base of $e$ to get $e^{ln(x)}.$ The $ln$ and $e$ cancel out, leaving you with $x$ .
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A log that is multiplied is the same as the sum of the logs. That might sound confusing, but this just means that when 2 terms in a log are multiplied, that’s the same as adding the log of the first term with the log of the second. Mathematically, this looks like $ln(ab)=ln(a)+ln(b).$ $ln(ab)=ln(a)+ln(b).$ This relationship makes it easy to group and combine logs with the same base, like natural logs, when there are multiple logs in an equation.^{[7]
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- Be careful of confusing $ln(ab)$ with $ln(a+b).$ $ln(a+b)$ does not equal $ln(a)+ln(b).$ ^{[8]
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3
A log that is divided is the same as the difference of the logs. This relationship between logs is similar to logs that are multiplied. When 2 terms of a log with the same base are divided, that’s like subtracting the log of the second term from the log of the first term. This relationship is written as $ln({\frac {a}{b}})=ln(a)-ln(b).$ $ln({\frac {a}{b}})=ln(a)-ln(b).$ ^{[9]
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- Be aware that $ln(a-b)$ is not the same as $ln(a)-ln(b)$ . The terms have to be divided by each other in order to equal $ln(a)-ln(b)$ .
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An exponential term in a log equals the exponent multiplied by the log. Some terms in a log have an exponent, which might look confusing at first. When you see them in an equation, that simply means that the exponent on the term in the log is the same as multiplying the exponent by the log. This is expressed like: $ln(a^{b})=b*ln(a)$ $ln(a^{b})=b*ln(a)$ .^{[10]
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- A log raised to an exponent does not follow the same properties as a log with a term raised to an exponent. For example, $(ln(a))^{b}$ does not equal $b*ln(a).$

Section 4 of 4:

Getting Rid of e

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$Step 1 Move all expressions with e {\displaystyle e} to 1 side of the equation.$

1
Move all expressions with $e$ to 1 side of the equation. The good news is that getting rid of $e$ $e$ follows a similar process as getting rid of $ln$ $ln$ . Start by isolating $e$ $e$ to 1 side of the equation.^{[11]
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- For example, solve the equation $200e^{0.07x}=500.$ $200e^{0.07x}=500.$
  - Divide each side by 200: $e^{0.07x}={\frac {500}{200}}.$
  - Simplify the fraction: $e^{0.07x}=2.5.$
$Step 2 Take the natural log of both sides of the equation to cancel out e {\displaystyle e} .$

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Take the natural log of both sides of the equation to cancel out $e$ . Because $e$ $e$ and $ln$ $ln$ are inverse functions, taking the natural log of $e$ $e$ gets rid of it. This brings down the exponent, leaving you with a solvable equation.^{[12]
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- Using the simplified equation $e^{0.07x}=2.5$ $e^{0.07x}=2.5$ :
  - Take the natural log of both sides of the equation: $ln(e^{0.07x})=ln(2.5).$
  - Cancel out $e$ and bring down its exponent: $0.07x=ln(2.5).$
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Solve the equation for x. Now that $e$ $e$ is out of the equation, simplify and solve the equation like normal. It’s likely that a natural log will still be in your equation, so follow your teacher’s instructions to either keep $ln$ $ln$ in the answer, or use a calculator to get a numerical answer for x.^{[13]
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- Solve $0.07x=ln(2.5).$ $0.07x=ln(2.5).$
  - Divide 0.07 by both sides: $x={\frac {ln(2.5)}{0.07}}$
  - Use a calculator to round the answer to the nearest thousandth place: $x=13.090.$

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Solve 2ln(x-5)= ln(x+7) +ln(6)

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First, using the ln property, we can bring the 2 back into the natural log. We get ln((x - 5)^2). Then, we can combine the two logs on the RHS to ln(6(x + 7). Then the ln's cancel, so (x - 5)^2 = 6(x + 7). Expanding, we get: x^2 - 10x + 25 = 6x + 42, so x^2 - 16x - 17 = 0. We can see that -17 + 1 = -16, so we can deduce that the factors are in the form (x - 17)(x + 1). The roots are x = 17 and x = -1.

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