Logarithm Calculator

A logarithm answers an exponent question. $\log_b x$ is the power you must raise $b$ to in order to get $x$:

$\log_b x = y \iff b^{\,y} = x$

So $\log_2 32 = 5$ because $2^5 = 32$, and $\log_{10} 1000 = 3$ because $10^3 = 1000$. Reading the definition aloud in that form turns most log questions into a fact you already know about powers.

Two bases have their own notation:

• $\log x$ with no base written means base 10 on virtually every calculator and in most school work.
• $\ln x$ means base $e \approx 2.71828$, the natural logarithm.

The domain is strictly positive: $x > 0$. There is no power of a positive base that produces 0 or a negative number, so $\log 0$ and $\log(-5)$ are undefined and your calculator will return an error, not a number.

Because of that inverse relationship, every logarithm question can be rewritten as a question about exponents, which is almost always the faster route by hand.

A calculator normally offers only $\log$ and $\ln$. For any other base, use change of base:

$\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}$

Either version works and both give the same value, because the constant factor between $\log$ and $\ln$ cancels in the quotient. That same relationship converts between them directly: $\ln x = \dfrac{\log x}{\log e} \approx 2.302585\,\log x$, and $\log x = \dfrac{\ln x}{\ln 10}$.

The three rules turn products into sums, which is what logarithms were invented for:

$\log_b(MN) = \log_b M + \log_b N$
$\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N$
$\log_b(M^{\,p}) = p\,\log_b M$

Plus two values worth memorising: $\log_b 1 = 0$ and $\log_b b = 1$.

To solve $\log_b x = c$, rewrite it in exponential form as $x = b^{\,c}$.

• Splitting a log over a sum. $\log(M + N)$ is not $\log M + \log N$. The sum rule applies to a product inside the log, never to an addition.
• Dividing instead of subtracting. $\dfrac{\log M}{\log N}$ is a change of base, not $\log\!\left(\frac{M}{N}\right)$. These are genuinely different numbers.
• Putting the base in the wrong place. Change of base is $\dfrac{\log(\text{argument})}{\log(\text{base})}$; flipping it gives the reciprocal.
• Assuming $\log$ means base $e$. On a calculator $\log$ is base 10; in some university texts it means base $e$. Check which convention your course uses.
• Taking the log of a negative number — undefined over the reals, so an equation step that produces one has produced an extraneous solution.
• Forgetting that $\log_b b^{\,x} = x$, which is often a one-line shortcut past a page of algebra.