Natural Logarithm

Natural logarithm finds the power to which e must be raised to get a number, and it's essential for solving equations involving exponential functions.

lnx=logex\ln x = \log_e x

Solve a problem with Natural Logarithm

Type the problem. The solver will use Natural Logarithm where Natural Logarithm is the right tool, and tell you when it is not.

Drag one in or paste from the clipboard. JPEG, PNG or WebP. You get the transcription to check before anything is solved.

How to get a better answer
  • Paste the whole problem, including the instruction word — "simplify", "solve for x" and "factor" lead to three different answers.
  • Say what you have already tried. "I got x = 4 and the book says 2" turns a solution into a diagnosis.
  • Set the level in the settings button. A calculus shortcut is not a better answer if you have not met derivatives yet.
  • For a photo, get the whole problem in frame and hold the page flat — you get the transcription to fix before anything is solved.

What each symbol means

What Natural Logarithm takes
xx
ee
Natural Logarithm
SymbolMeaning
xxThe positive input value to the logarithm; xx must be strictly greater than zero or lnx\ln x is undefined.
eeEuler's number, approximately 2.71828, an irrational constant that appears in exponential growth and is the base of natural logarithm.

When to use it

Reach for natural logarithm when you need to undo an exponential equation with base e or solve for a variable in the exponent.

Level

Usually taught in: Algebra II

Worked examples

1. Evaluate using the Power Rule

Problem

Evaluate ln(e4)\ln(e^4).
  1. ln(e4)=4ln(e)\ln(e^4) = 4 \ln(e)

    Apply the Power Rule for Logarithms: ln(ab)=bln(a)\ln(a^b) = b \ln(a).

  2. ln(e)=1\ln(e) = 1

    The logarithm of a number to its own base is always 1, since e1=ee^1 = e.

  3. 41=44 \cdot 1 = 4

    Multiply to simplify.

Answer: ln(e4)=4\ln(e^4) = 4

The Power Rule for Logarithms lets us bring the exponent down in front. Since ln(e)=1\ln(e) = 1 by definition, the final answer is just the exponent itself.

2. Solve an exponential equation

Problem

Solve ln(x)=3\ln(x) = 3 for xx.
  1. ln(x)=3\ln(x) = 3

    This is the given equation in logarithmic form.

  2. x=e3x = e^3

    Convert from logarithmic to exponential form: if ln(x)=3\ln(x) = 3, then e3=xe^3 = x.

  3. x20.09x \approx 20.09

    Evaluate e3e^3 using a calculator: e320.0855e^3 \approx 20.0855.

Answer: x=e3 or x20.09x = e^3 \text{ or } x \approx 20.09

Natural logarithm and the exponential function with base ee are inverse operations. To solve ln(x)=3\ln(x) = 3, convert to exponential form by raising ee to both sides.

3. Real-world exponential model in basketball

Problem

A basketball player's cumulative scoring during a game follows P(t)=10e0.2tP(t) = 10e^{0.2t}, where PP is cumulative points and tt is time in minutes. After how many minutes do they reach 30 points?
  1. 10e0.2t=3010e^{0.2t} = 30

    Set up the equation by substituting P=30P = 30 into the model.

  2. e0.2t=3e^{0.2t} = 3

    Divide both sides by 10.

  3. ln(e0.2t)=ln(3)\ln(e^{0.2t}) = \ln(3)

    Take the natural logarithm of both sides to undo the exponential.

  4. 0.2t=ln(3)0.2t = \ln(3)

    Use the Power Rule: ln(e0.2t)=0.2t\ln(e^{0.2t}) = 0.2t, and ln(3)1.0986\ln(3) \approx 1.0986.

  5. t=ln(3)0.2t = \frac{\ln(3)}{0.2}

    Divide both sides by 0.2 to isolate tt.

  6. t5.49t \approx 5.49

    Calculate: 1.09860.25.493\frac{1.0986}{0.2} \approx 5.493 minutes.

Answer: t=ln(3)0.25.49 minutest = \frac{\ln(3)}{0.2} \approx 5.49 \text{ minutes}

Real-world exponential growth and decay models often require natural logarithm to solve. Taking ln\ln of both sides converts an exponential equation into an algebraic one, allowing us to isolate the exponent and solve for the variable.

Common mistakes

Where Natural Logarithm usually goes wrong
Answer came out wrong
ln(x+y)=ln(x)+ln(y)\ln(x + y) = \ln(x) + \ln(y)
The Product Rule is ln(xy)=ln(x)+ln(y)\ln(xy) = \ln(x) + \ln(y); for sums like ln(x+y)\ln(x + y), you must evaluate the logarithm directly without breaking it apart.
Concluding that ln(x)=2\ln(x) = -2 has no solution because x=e2x = e^{-2} seems too small.
x=e2=1e20.135x = e^{-2} = \frac{1}{e^2} \approx 0.135 is the correct answer; negative logarithmic outputs are valid when the input is between 0 and 1.
Attempting to evaluate ln(5)\ln(-5) or ln(0)\ln(0) as real numbers.
Always verify that the input to ln\ln is strictly positive; if it is not, the equation has no real solution.
The mistakeWhy it is wrongThe fix
ln(x+y)=ln(x)+ln(y)\ln(x + y) = \ln(x) + \ln(y)This incorrectly treats addition as if it were multiplication; the Product Rule only applies to products inside the logarithm, not sums.The Product Rule is ln(xy)=ln(x)+ln(y)\ln(xy) = \ln(x) + \ln(y); for sums like ln(x+y)\ln(x + y), you must evaluate the logarithm directly without breaking it apart.
Concluding that ln(x)=2\ln(x) = -2 has no solution because x=e2x = e^{-2} seems too small.Natural logarithm of numbers between 0 and 1 yields negative results, which is correct and expected behavior.x=e2=1e20.135x = e^{-2} = \frac{1}{e^2} \approx 0.135 is the correct answer; negative logarithmic outputs are valid when the input is between 0 and 1.
Attempting to evaluate ln(5)\ln(-5) or ln(0)\ln(0) as real numbers.Natural logarithm is undefined for zero and negative numbers; there is no real exponent to which you can raise ee and get a negative or zero result.Always verify that the input to ln\ln is strictly positive; if it is not, the equation has no real solution.

Tips and when to use something else

  • Natural logarithm and exponential functions with base ee are inverses: ln(ex)=x\ln(e^x) = x and eln(x)=xe^{\ln(x)} = x for all x>0x > 0.
  • When solving exponential equations like ex=e^x = a number, use natural logarithm; for other bases, use Change of Base Formula to convert to base ee first.
  • Natural logarithm appears constantly in science and engineering—radioactive decay, compound interest, cooling rates, population growth—because ee is how nature naturally grows.
  • Sanity check: ln(1)=0\ln(1) = 0 always, since e0=1e^0 = 1; use this to verify your mental math and catch errors.

Frequently asked questions

Why is it called the 'natural' logarithm?
It is called natural because Euler's number ee appears naturally throughout mathematics, physics, and biology—in growth, decay, compound interest, and many other phenomena. When mathematicians studied these natural processes, the logarithm with base ee kept emerging as the most useful, so it earned the name natural logarithm.
How is natural logarithm different from log base 10?
Natural logarithm uses base e2.718e \approx 2.718, while common logarithm (written as log\log or log10\log_{10}) uses base 10. Natural logarithm is used in calculus, exponential modeling, and science; common logarithm is useful for counting significant figures and working with scientific notation.
What does ln(e)\ln(e) equal and why?
ln(e)=1\ln(e) = 1 because natural logarithm asks 'what power must ee be raised to in order to produce ee?', and the answer is 1, since e1=ee^1 = e. More generally, logb(b)=1\log_b(b) = 1 for any valid logarithm base.
Can natural logarithm be negative?
Yes, natural logarithm is negative when the input is between 0 and 1. For example, ln(0.5)0.693\ln(0.5) \approx -0.693 because e0.6930.5e^{-0.693} \approx 0.5. However, you cannot take the natural logarithm of a negative number or zero—these are undefined in the real numbers.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18