Mathematics RU

Powers and logarithms

The laws of exponents, values of logarithms, their laws and simple equations like 4ˣ = 2. Problem after problem, fully explained.

How to solve it

A logarithm is an exponent: $\log_a b$ answers the question to what power $a$ must be raised to get $b$. So the laws of logarithms are the laws of exponents read backwards: when powers are multiplied the exponents add, so the logarithm of a product is the sum of the logarithms.

Step by step

Collect powers of one base into one: multiplying adds the exponents, dividing subtracts them, raising to a power multiplies them.
The value of a logarithm: write the number as a power of the base, $\frac{1}{16} = 2^{-4}$, so $\log_2 \frac{1}{16} = -4$.
If the base and the number are powers of the same number, rewrite both: $\log_4 32$: $4 = 2^2$, $32 = 2^5$, the answer is $\frac{5}{2}$.
Replace a sum of logarithms with one base by the logarithm of the product: $\lg 4 + \lg 250 = \lg 1000 = 3$.
An exponential equation: bring both sides to one base and equate the exponents.

The base: $a > 0$, $a \ne 1$.

The number under the logarithm, always positive.

The logarithm itself, an exponent.

Example: $4^x = 2$; $4 = 2^2$, so $2^{2x} = 2^1$, $2x = 1$, $x = \frac{1}{2}$.

Common mistakes

Multiplying the exponents when multiplying powers: $5^6 \cdot 5^{-3} = 5^3$, not $5^{-18}$.
Confusing a negative exponent with a negative number: $10^{-1} = \frac{1}{10}$, not $-10$.
Thinking the logarithm of a sum is the sum of the logarithms. No: it is the logarithm of a product, $\lg(a + b) \ne \lg a + \lg b$.
Forgetting that $\lg$ is the logarithm to base $10$ and $\ln$ to base $e$.