Exponents, Exponentials and Logarithms · Lesson 3

What a logarithm means

A logarithm is an exponent. log⁡bx=y\log_b x = y means by=xb^y = x. So log⁡232\log_2 32 asks: 2 to what power gives 32? The answer is 5, because 25=322^5 = 32. Read every log as that question.

Two bases have their own notation. log⁡x\log x with no base written is the common log, base 10, so log⁡1,000=3\log 1{,}000 = 3. ln⁡x\ln x is the natural log, base ee (about 2.718), so ln⁡ek=k\ln e^k = k.

Special values. log⁡b1=0\log_b 1 = 0 because b0=1b^0 = 1, and log⁡bb=1\log_b b = 1 because b1=bb^1 = b. A fraction less than 1 gives a negative log: log⁡218=−3\log_2 \frac{1}{8} = -3 because 2−3=182^{-3} = \frac{1}{8}. When the argument isn't a simple power of the base, rewrite both with a shared base: log⁡927=y\log_9 27 = y means 32y=333^{2y} = 3^3, so y=32y = \frac{3}{2}.

Domain. The input of a log must be positive. log⁡b0\log_b 0 and the log of a negative number are undefined, so log⁡(x−4)\log(x - 4) needs x>4x > 4.

Worked example Easy

What is the value of log⁡381\log_3 81?

  1. 14\frac{1}{4}
  2. 44Answer
  3. 99
  4. 2727

How to solve it

  1. log⁡381\log_3 81 asks: 3 to what power gives 81?
  2. 31=33^1 = 3, 32=93^2 = 9, 33=273^3 = 27, 34=813^4 = 81.
  3. So log⁡381=4\log_3 81 = 4.

Why each choice is right or wrong

  • A. Incorrect. 14\frac{1}{4} is log⁡813\log_{81} 3, with the base and the argument swapped. The small number written low is the base, so ask what power of 3 gives 81.
  • B. Correct. 34=813^4 = 81, so log⁡381=4\log_3 81 = 4.
  • C. Incorrect. 9 is the square root of 81. A log asks for an exponent: 3 must be raised to the 4th power to reach 81.
  • D. Incorrect. 27 divides 81 by 3. A log counts how many factors of 3 make 81, and 81=3⋅3⋅3⋅381 = 3 \cdot 3 \cdot 3 \cdot 3 has four.

Worked example Medium

What is the value of log⁡432\log_4 32?

  1. 25\frac{2}{5}
  2. 52\frac{5}{2}Answer
  3. 55
  4. 88

How to solve it

  1. Write log⁡432=y\log_4 32 = y, so 4y=324^y = 32.
  2. Rewrite both sides with base 2: (22)y=25(2^2)^y = 2^5, so 22y=252^{2y} = 2^5.
  3. Set the exponents equal: 2y=52y = 5, so y=52y = \frac{5}{2}.
  4. Check: 45/2=(4)5=25=324^{5/2} = (\sqrt{4})^5 = 2^5 = 32.

Why each choice is right or wrong

  • A. Incorrect. 25\frac{2}{5} is log⁡324\log_{32} 4, which swaps the base and the argument. Solve 4y=324^y = 32 instead.
  • B. Correct. 4y=324^y = 32 becomes 22y=252^{2y} = 2^5, so 2y=52y = 5 and y=52y = \frac{5}{2}.
  • C. Incorrect. 5 is log⁡232\log_2 32. The base here is 4, and since 4=224 = 2^2, the exponent is half as large.
  • D. Incorrect. 8 divides 32 by 4. A log asks for an exponent, not a quotient.

Trap Swapping the base and the argument

log⁡381\log_3 81 is 4, but log⁡813\log_{81} 3 is 14\frac{1}{4}. The subscript is always the base, the number that gets raised to a power. When you rewrite a log as a power, the base stays the base: log⁡bx=y\log_b x = y becomes by=xb^y = x.

Trap $\log_b 1$ is 0, not 1

Any nonzero base to the power 0 is 1, so log⁡b1=0\log_b 1 = 0. It is log⁡bb\log_b b that equals 1. A quick test: log⁡1=0\log 1 = 0 because 100=110^0 = 1.

Calculator When a calculator isn’t faster

For logs of exact powers, mental work wins: ask what power of the base gives the argument. Your calculator has log (base 10) and ln keys, and the built-in Desmos calculator also accepts a base, as in log_4(32).

On a scientific calculator without a base key, use log⁡32log⁡4\frac{\log 32}{\log 4}. A decimal answer such as 2.5 still has to be matched to a choice like 52\frac{5}{2}.