Exponents, Logarithms and nth Roots, Explained
After reading this you can compute a logarithm to any base, solve b^x = y for the unknown exponent, and take exact nth roots without second-guessing which operation you need.
What these three operations share
Exponentiation, logarithms and roots are three ways of looking at the same relationship. Start with the equation b^x = y. It ties together three numbers: a base b, an exponent x and a result y. Fix any two and you can find the third.
If you know b and x, you exponentiate: 2^3 = 8. If you know b and y, you take a logarithm to recover the exponent: \log_2 8 = 3. If you know x and y, you take a root to recover the base: \sqrt[3]{8} = 2.
So the logarithm mode and the root mode both invert exponentiation. They just solve for different unknowns. Getting this straight is the whole battle. Once you know which quantity is missing, the operation picks itself.
When to use each mode
Use the logarithm mode when you have the base and the result and want the exponent. Questions like "2 to what power gives 8?" or "how many times must I multiply by 1.05 to double?" are logarithm questions.
Use the solve b^x = y mode for the same idea stated as an equation. It is the natural choice when a problem hands you an exponential equation directly, such as 3^x = 20.
Use the nth root mode when you know the exponent and the result and want the base. Average growth rates work this way. If money grows from 1000 to 1610 over 5 years, the yearly factor is \sqrt[5]{1.61} \approx 1.10, so about 10 percent a year.
Logarithms are defined only for a positive value and a positive base other than 1. You cannot compute \log_2(-4) or use base 1, because 1^x = 1 for every x and can never equal any other number. For roots of negative numbers, see the FAQ.
The formulas and the intuition
The defining statement is the equivalence between an exponential form and a logarithmic form.
Here b is the base, y is the value you feed in, and x is the exponent the logarithm returns. Read \log_b(y) as "the power you raise b to in order to get y."
Most calculators store only natural log (base e) and base-10 log. To reach any base, use the change of base rule.
You can put \log_{10} on top and bottom instead. The ratio is the same because both numerators and denominators scale by the same constant. For example, \log_2 8 = \ln 8 / \ln 2 = 2.0794 / 0.6931 = 3.
The nth root is exponentiation by a fraction.
Here n is the root order (a whole number like 2 for square root, 3 for cube root) and y is the value. Cubing and cube-rooting cancel: (y^{1/3})^3 = y^{3/3} = y.
A worked example: log base 2 of 8
Reproducing the demo
The demo button loads mode = log, base = 2, value = 8. You are asking: 2 to what power equals 8?
- Write the question as an equation: 2^x = 8.
- Recognize that 8 = 2 \cdot 2 \cdot 2 = 2^3.
- Match exponents: since 2^x = 2^3, you get x = 3.
- Check with change of base: \ln 8 / \ln 2 = 2.0794 / 0.6931 = 3.000.
The answer is exactly 3. Now change the value to 10 with the base still 2. There is no clean power of 2 equal to 10, so you get \log_2 10 = 3.3219. That says 10 sits between 2^3 = 8 and 2^4 = 16, closer to the low end, which matches 3.3219 being just above 3.
Watch the shape of the log curve
A logarithm grows fast at first and then crawls. Doubling the input adds a fixed amount to the output, not a fixed percentage. For base 2, every doubling of y raises \log_2 y by exactly 1: \log_2 4 = 2, \log_2 8 = 3, \log_2 16 = 4.
Reading and interpreting the result
A logarithm is a count of multiplications. \log_2 8 = 3 means "three doublings from 1 reach 8." A base-10 log answers "how many digits, roughly?": \log_{10} 5000 = 3.699, and 5000 has 4 digits, consistent with a log between 3 and 4.
Signs carry meaning. When the value is less than 1, the logarithm is negative: \log_2 0.25 = -2, because 2^{-2} = 0.25. When the value equals 1, the logarithm is 0 for every valid base, since b^0 = 1.
For roots, interpret the result as the number that, raised to the power n, gives back the value. \sqrt[4]{81} = 3 because 3^4 = 81. If the root is not exact, the display shows a decimal you can square or cube to check.
| Value y | base 2 | base e | base 10 |
|---|---|---|---|
| 2 | 1 | 0.6931 | 0.3010 |
| 8 | 3 | 2.079 | 0.9031 |
| 100 | 6.644 | 4.605 | 2 |
| 0.5 | -1 | -0.6931 | -0.3010 |
Common mistakes
The most frequent slip is confusing the base and the value. \log_2 8 = 3 but \log_8 2 = 1/3. Swapping them flips the answer to its reciprocal, so always ask which number is the base you are raising.
The logarithm of a sum is not the sum of the logarithms. \log_2(4 + 4) = \log_2 8 = 3, but \log_2 4 + \log_2 4 = 2 + 2 = 4. The product rule \log(ab) = \log a + \log b applies to products, never to sums.
Another trap is an even root of a negative number. \sqrt{-9} has no real value, because no real number squared is negative. Odd roots are fine: \sqrt[3]{-8} = -2 since (-2)^3 = -8. For even roots of negatives you need the Complex number calculator.
Finally, do not confuse \ln (base e \approx 2.718) with \log written without a base. In many textbooks a bare \log means base 10, in others it means base e. State the base you mean.
Related tools
If your equation involves polynomials rather than exponents, reach for the Equation solver or the Quadratic equation solver. To simplify a root into exact surd form like \sqrt{72} = 6\sqrt{2}, use the Radical simplifier. Exponential and geometric patterns across a list of terms belong to the Sequence calculator, which handles nth terms and sums directly.
Frequently asked questions
How do I compute a log to a base my calculator does not have?
Divide two logs in a base you do have. \log_5 20 = \ln 20 / \ln 5 = 2.996 / 1.609 = 1.861. Using base-10 logs gives the same result: 1.301 / 0.6990 = 1.861.
What is the difference between solving bˣ = y and taking a log?
There is no mathematical difference. Solving b^x = y for x is exactly evaluating \log_b y. The solve mode is a convenience for when the problem is already written as an equation.
Why is log of 1 always zero?
Because any nonzero base raised to the power 0 equals 1: b^0 = 1. Reading the equivalence backward, \log_b 1 = 0 for every valid base b.
Can I take a fractional or negative root order?
Yes, through the exponent form. A negative order inverts: y^{-1/2} = 1/\sqrt{y}, so an order of -2 on 100 gives 1/10 = 0.1. A fractional order like y^{2/3} combines a cube root and a square.
How many digits does a big number have?
Take the base-10 log, drop the fractional part, and add 1. \log_{10} 987654 = 5.9946, floor is 5, plus 1 gives 6 digits, which matches.
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