How to Calculate a Logarithm Without a Calculator

A logarithm asks which exponent produces a given number: $\log_b x=y$ means $b^y=x$. Some logarithms can be found exactly by recognizing powers. For other values, you can simplify and estimate by hand, or use a series to obtain as many decimal places as needed.

The method depends on whether you need an exact answer, a rough estimate, or a numerical approximation. There is no general rule that turns every logarithm into a simple exact number.

Look for exact logarithms first

If the input is a power of the base, use the definition directly. For example,

$$\log_2 32=5$$

because $2^5=32$. Likewise, $\log_3(1/27)=-3$, since $3^{-3}=1/27$. A negative logarithm is expected when the input is between zero and one and the base is greater than one.

Logarithm laws can reveal powers that are not obvious at first:

$$\log_b(xy)=\log_b x+\log_b y,\qquad \log_b\left(\frac{x}{y}\right)=\log_b x-\log_b y,\qquad \log_b(x^r)=r\log_b x.$$

These rules apply when the logarithm arguments are positive and the base satisfies $b>0$, $b\ne1$. For instance, $\log_2 40=\log_2(8\cdot5)=3+\log_2 5$. That is an exact simplification, even though it does not by itself give a decimal.

Estimate using nearby powers

For a base $b>1$, find consecutive powers that enclose the input. If $b^n\le x<b^{n+1}$, then

$$n\le\log_b x<n+1.$$

For example, since $2^3=8<10<16=2^4$, you know that $\log_2 10$ lies between 3 and 4. To narrow the estimate, compare $10$ with powers having fractional exponents if you know them, or use a numerical method.

It can help to split the input into a convenient power times a smaller factor. For example, $\log_{10} 1200=2+\log_{10}12$, because $1200=100\cdot12$. This does not always finish the calculation, but it can make the remaining logarithm easier to estimate or look up in a table.

Use change of base

When you can approximate natural logarithms, convert any positive-base logarithm using

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

This identity is valid for $x>0$, $b>0$, and $b\ne1$. It is useful because there are practical series for $\ln x$. If your class or reference gives a table of common logarithms instead, the same idea works with base 10: $\log_b x=\log_{10}x/\log_{10}b$.

Approximate a natural logarithm with a series

For $x>0$, define

$$u=\frac{x-1}{x+1}.$$

Then

$$\ln x=2\left(u+\frac{u^3}{3}+\frac{u^5}{5}+\frac{u^7}{7}+\cdots\right).$$

This series comes from integrating the geometric series, and it converges for every positive $x$ because $|u|<1$. It converges especially quickly when $x$ is close to 1. If $x$ is far from 1, first factor out a convenient power: for example, $\ln(8x)=3\ln2+\ln x$.

Worked example: approximate $\ln(1.2)$

Here,

$$u=\frac{1.2-1}{1.2+1}=\frac{0.2}{2.2}=\frac{1}{11}.$$

Keeping the first three terms gives

$$\begin{aligned} \ln(1.2)&\approx 2\left(\frac{1}{11}+\frac{1}{3\cdot11^3}+\frac{1}{5\cdot11^5}\right)\\ &\approx 2(0.090909091+0.000250438+0.000001242)\\ &\approx 0.182321542. \end{aligned}$$

The next omitted term is $2/(7\cdot11^7)$, about $0.0000000147$, and the remaining tail is only slightly larger. The truncation error is about $0.0000000148$, so the approximation is within $0.00000002$ of the true value; to six decimal places, $\ln(1.2)\approx0.182322$.

More generally, if you keep terms through $u^{2N+1}/(2N+1)$, the absolute size of the remaining tail is at most

$$\frac{2|u|^{2N+3}}{(2N+3)(1-u^2)}.$$

This bound lets you decide how many terms are sufficient instead of guessing.

Common mistakes

• Mixing up bases: $\ln x$ means $\log_e x$, while $\log x$ often means base 10 in school contexts. Check the convention being used.
• Applying product rules to sums: In general, $\log_b(x+y)$ is not $\log_b x+\log_b y$.
• Ignoring the input’s domain: Real logarithms require a positive input. For real-valued logarithms, the base must be positive and not equal to 1.
• Rounding too early: Keep extra digits during intermediate calculations, especially when dividing two approximated logarithms in the change-of-base formula.
• Expecting a simple exact answer: A logarithm such as $\log_{10}2$ generally needs to be left in logarithmic form or approximated; it is not a simple rational number.

For a hand calculation, start with exact power recognition and log laws. Use nearby powers for a quick range, and use the series when you need a controlled decimal approximation.