Natural Log (ln) Calculator

The natural logarithm is the logarithm with base $e \approx 2.718281828$:

$\ln x = \log_e x, \qquad \ln x = y \iff e^{\,y} = x$

So $\ln e = 1$, $\ln 1 = 0$, and $\ln e^{5} = 5$. Nothing about the rules is special — $e$ is simply an unusually convenient base, because the function $e^{x}$ is its own derivative. That is why continuous growth, radioactive decay, compound interest and half-lives all come out in $\ln$ rather than $\log$.

The two functions are inverses, and that is the property you use constantly:

$\ln(e^{x}) = x \qquad \text{and} \qquad e^{\ln x} = x \quad (x > 0)$

Applying $\ln$ to both sides of an equation is how you pull a variable down out of an exponent — the single most common reason to reach for it.

The domain is $x > 0$: $\ln 0$ and $\ln(-2)$ do not exist over the real numbers.

On a calculator the $e^{x}$ function normally sits directly above the $\ln$ key, which is a useful physical reminder that the two undo each other.

Variable in the exponent. Take $\ln$ of both sides and use $\ln(e^{x}) = x$:

$e^{2x} = 15 \quad \Rightarrow \quad 2x = \ln 15 \quad \Rightarrow \quad x = \tfrac{1}{2}\ln 15$

For a base that is not $e$, the power rule does the same job: $3^{x} = 20$ becomes $x\ln 3 = \ln 20$.

Variable inside a log. Exponentiate both sides — apply $e$ to undo $\ln$:

$\ln(x - 3) = 2 \quad \Rightarrow \quad x - 3 = e^{2} \quad \Rightarrow \quad x = 3 + e^{2}$

Several logs. Combine them into a single one with the product, quotient and power rules, then drop the logs from both sides: if $\ln A = \ln B$ then $A = B$.

Always check the domain at the end. Every solution must keep every original log argument strictly positive. Combining logs can manufacture solutions the original equation never had.

• Skipping the domain check. Combining $\ln x + \ln(x-3)$ produces a quadratic with two roots, and typically one of them makes an argument negative and must be rejected.
• Writing $\ln(A + B) = \ln A + \ln B$. The sum rule works on a product inside the log, never on an addition.
• Cancelling the ln from only one side. From $\ln x = 2$ you get $x = e^{2}$, not $x = 2$.
• Confusing $\ln x^{2}$ with $(\ln x)^{2}$. The first equals $2\ln x$. The second is a square, and the power rule does not touch it.
• Treating $\dfrac{\ln 20}{\ln 3}$ as $\ln\!\left(\frac{20}{3}\right)$. The first is $\log_3 20 \approx 2.727$. The second is about $1.897$.
• Reaching for $\log$ when the base is $e$. Either works if you stay consistent, but $\ln$ removes an $e$ in one step.