Natural Logarithms (ln)

Learn what ln x means, why ln e = 1 and ln 1 = 0, how ln undoes e^x, and how to solve equations like e^x = 20 with a calculator.

Advanced25 min18 problems
Practice Problems

Definition

The natural logarithm is the logarithm with base ee, where e≈2.71828e \approx 2.71828 is Euler's number. Instead of writing log⁡ex\log_e x, we write ln⁡x\ln x (read "el-en of x" or "natural log of x").
ln⁡x=log⁡ex\ln x = \log_e x
Like every logarithm, ln⁡x\ln x answers a question: **"To which power must I raise ee to get xx?"**
ln⁡x=y  ⟺  ey=x\ln x = y \iff e^y = x
Two values to know by heart:
  • ln⁡1=0\ln 1 = 0, because e0=1e^0 = 1
  • ln⁡e=1\ln e = 1, because e1=ee^1 = e
ln undoes e^x (and the other way round): the function y=ln⁡xy = \ln x is the inverse function of y=exy = e^x, so
ln⁡(ex)=xandeln⁡x=x  (x>0)\ln(e^x) = x \quad \text{and} \quad e^{\ln x} = x \; (x > 0)
The argument must be positive: ln⁡x\ln x only exists for x>0x > 0, because eye^y is never zero or negative. So the domain of y=ln⁡xy = \ln x is x>0x > 0.

Try it now

What is the value of ln⁡1\ln 1?

Worked Examples

Evaluate ln⁡(e5)\ln(e^5) without a calculator.
1

Rewrite as a question

ln⁡(e5)\ln(e^5) asks: to which power must I raise ee to get e5e^5? → e?=e5e^{?} = e^5
2

Read off the exponent

The power that turns ee into e5e^5 is 55 → ?=5? = 5
3

Use the inverse rule

ln⁡\ln undoes exe^x: ln⁡(ex)=x\ln(e^x) = x → ln⁡(e5)=5\ln(e^5) = 5

Common Mistakes

Writing ln⁡1=1\ln 1 = 1 or ln⁡e=0\ln e = 0

Why it's wrong: The two key values are easy to swap. ln⁡\ln gives an exponent, not the number itself.

Correct: ln⁡1=0\ln 1 = 0 because e0=1e^0 = 1, and ln⁡e=1\ln e = 1 because e1=ee^1 = e.

Taking ln⁡\ln of zero or a negative number, like ln⁡0\ln 0 or ln⁡(−3)\ln(-3)

Why it's wrong: eye^y is always positive, whatever yy is. No power of ee gives 00 or a negative number.

Correct: ln⁡x\ln x only exists for x>0x > 0. The calculator shows an error for ln⁡0\ln 0 and ln⁡(−3)\ln(-3).

Pressing LOG instead of LN on the calculator

Why it's wrong: LOG is the common logarithm with base 10, LN is the natural logarithm with base ee. They give different results.

Correct: log⁡10=1\log 10 = 1, but ln⁡10≈2.303\ln 10 \approx 2.303. Use LN whenever the base is ee.

Solving ex=20e^x = 20 by dividing: x=20ex = \frac{20}{e}

Why it's wrong: xx is an exponent, not a factor. Dividing by ee does not undo raising ee to a power.

Correct: Undo exe^x with ln⁡\ln: x=ln⁡20≈2.996x = \ln 20 \approx 2.996.

Interactive Visual

Function graph

-4-202468-4-202468xy(0, 1)(1, 0)(1, 2.72)(2.72, 1)y = xy = 0x = 0

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y = xAsymptote

Table of values

x0.51235
e^x1.648722.718287.3890620.0855148.413
ln(x)-0.69314700.6931471.098611.60944

Interactive Sandbox

Interactive Grapher

-10-50510-10-50510xy

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y = 2x + 1

m=2, b=1

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Practice Problems

18 problems
Problem 1 of 18
Easy
What is the value of ln⁡1\ln 1?

Why It Matters

The number ee appears whenever something grows or shrinks continuously, and ln⁡\ln is the tool that brings the time or the rate back out of the exponent:
  • Continuous growth: if bacteria grow as N=500e0.4tN = 500e^{0.4t}, then ln⁡\ln tells you when there will be 2000 of them.
  • Money: with continuous compounding A=PertA = Pe^{rt}, the doubling time is t=ln⁡2rt = \frac{\ln 2}{r}.
  • Medicine: how long a drug stays above a safe level in the blood is found with ln⁡\ln.
  • Science and calculus: ln⁡x\ln x shows up in physics, chemistry and every calculus course, because exe^x and ln⁡x\ln x have the simplest derivatives.
Without ln⁡\ln, an equation like ex=20e^x = 20 could only be solved by guessing. With ln⁡\ln, it takes one step: x=ln⁡20≈2.996x = \ln 20 \approx 2.996.

Real World Applications

Biology: When Will the Culture Reach a Target?

A bacteria culture with plenty of food grows continuously as $N = N_0 e^{rt}$. To find WHEN it reaches a size, you need $\ln$ to bring $t$ down from the exponent.

Example:

With N=1000e0.3tN = 1000e^{0.3t}, the culture reaches 3000 when e0.3t=3e^{0.3t} = 3, so 0.3t=ln⁡30.3t = \ln 3 and t=ln⁡30.3≈3.66t = \frac{\ln 3}{0.3} \approx 3.66 hours.

1Try It Yourself

A culture starts with 500 bacteria and grows as N=500e0.4tN = 500e^{0.4t}, where tt is in hours.

After how many hours are there 2000 bacteria? Round to 2 decimal places.

Step 1: Write the mathematical expression

Calculate t=ln⁡40.4t = \frac{\ln 4}{0.4}:

Finance: Doubling Time with Continuous Interest

Banks compare savings offers with continuous compounding, $A = Pe^{rt}$. The time it takes to double money is $t = \frac{\ln 2}{r}$.

Example:

At 7% compounded continuously, money doubles after ln⁡20.07≈9.90\frac{\ln 2}{0.07} \approx 9.90 years. Since ln⁡2≈0.693\ln 2 \approx 0.693, bankers use the "rule of 70": 70 divided by the rate in percent.

2Try It Yourself

You invest 1000 dollars at 5% interest compounded continuously, so A=1000e0.05tA = 1000e^{0.05t}.

After how many years will you have 2000 dollars? Round to 2 decimal places.

Step 1: Write the mathematical expression

Calculate t=ln⁡20.05t = \frac{\ln 2}{0.05}:

Medicine: How Long a Drug Stays in the Blood

Many medicines leave the body continuously, so the amount in the blood follows $C = C_0 e^{-kt}$. Doctors use $\ln$ to work out when the next dose is due.

Example:

A drug with k=0.25k = 0.25 per hour has a half-life of ln⁡20.25≈2.77\frac{\ln 2}{0.25} \approx 2.77 hours: after that time, half of it is left.

3Try It Yourself

A patient takes 100 mg of a medicine. The amount left after tt hours is C=100e−0.25tC = 100e^{-0.25t}.

After how many hours are 40 mg left? Round to 2 decimal places.

Step 1: Write the mathematical expression

Calculate t=ln⁡0.4−0.25t = \frac{\ln 0.4}{-0.25}:

Key Takeaways

  • 1The natural logarithm is the logarithm with base e≈2.71828e \approx 2.71828: ln⁡x=log⁡ex\ln x = \log_e x
  • 2ln⁡x=y\ln x = y means ey=xe^y = x: ln gives the power of ee that produces xx
  • 3Key values: ln⁡1=0\ln 1 = 0 and ln⁡e=1\ln e = 1
  • 4ln⁡\ln and exe^x are inverse functions: ln⁡(ex)=x\ln(e^x) = x and eln⁡x=xe^{\ln x} = x for x>0x > 0
  • 5ln⁡x\ln x only exists for x>0x > 0; it is negative for 0<x<10 < x < 1 and positive for x>1x > 1
  • 6To solve ex=ke^x = k (with k>0k > 0), take ln of both sides: x=ln⁡kx = \ln k

Frequently Asked Questions

Because its base ee comes up naturally: in continuous growth, in continuous compounding and in calculus, where exe^x and ln⁡x\ln x have the simplest rates of change. The "n" in ln⁡\ln stands for the Latin "logarithmus naturalis".
Because its base ee comes up naturally: in continuous growth, in continuous compounding and in calculus, where exe^x and ln⁡x\ln x have the simplest rates of change. The "n" in ln⁡\ln stands for the Latin "logarithmus naturalis".
No. log⁡x\log x (common logarithm) has base 10 and ln⁡x\ln x has base ee. For example, log⁡100=2\log 100 = 2 but ln⁡100≈4.605\ln 100 \approx 4.605. Both answer the same kind of question, only the base is different.
Because eye^y is always positive: there is no power of ee that equals 00 or a negative number. The graph of y=ln⁡xy = \ln x only lives to the right of the yy-axis, which is a vertical asymptote.
They are mirror images of each other in the line y=xy = x, because they are inverse functions. The point (0,1)(0, 1) on y=exy = e^x matches the point (1,0)(1, 0) on y=ln⁡xy = \ln x, and (1,e)(1, e) matches (e,1)(e, 1).

Glossary

Natural logarithm
The logarithm with base ee, written ln⁡x\ln x; ln⁡x=y\ln x = y means ey=xe^y = x
Euler's number ee
The irrational number e≈2.71828e \approx 2.71828, the base of continuous growth
Inverse function
A function that undoes another one; ln⁡x\ln x and exe^x are inverse functions
Argument
The number you take the logarithm of, the xx in ln⁡x\ln x; it must be positive
Domain
All allowed inputs of a function; for y=ln⁡xy = \ln x the domain is x>0x > 0
Exponential equation
An equation with the unknown in the exponent, such as ex=20e^x = 20

Formula Card

Definition

ln⁡x=log⁡ex\ln x = \log_e x

Logarithm with base $e \approx 2.71828$

Exponential form

ln⁡x=y  ⟺  ey=x\ln x = y \iff e^y = x

Switch between the two forms

Key values

ln⁡1=0,ln⁡e=1\ln 1 = 0, \quad \ln e = 1

Because $e^0 = 1$ and $e^1 = e$

Inverse rules

ln⁡(ex)=x,eln⁡x=x\ln(e^x) = x, \quad e^{\ln x} = x

The second one for $x > 0$

Solving $e^x = k$

x=ln⁡kx = \ln k

Only for $k > 0$

Today's challengeOne problem, three tries

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