Exponentials and Logarithms

The graph of $y=2^x$y = 2ˣ

In $y=2^x$y = 2ˣ the variable x is the exponent. Work out some values. At x = 0, $y=2^0=1$y = 2⁰ = 1. At x = 1, 2, 3, 4 and 5, y is 2, 4, 8, 16 and 32. Each step of 1 to the right multiplies y by 2.

To the left the values halve instead: $2^{-1}=\frac{1}{2}$2⁻¹ = 1/2, $2^{-2}=\frac{1}{4}$2⁻² = 1/4 and $2^{-3}=\frac{1}{8}$2⁻³ = 1/8, because a negative index means one over the power. The values get closer and closer to 0, but they never reach it, because half of a positive number is still positive.

So on the left the graph stays just above the x-axis. It passes through (0, 1) and then climbs faster and faster. Between x = 4 and x = 5 it rises by 16, which is more than its whole rise from x = 0 to x = 4, 16 − 1 = 15.

$y=2^x$y = 2ˣ. The marked points are (0, 1), (1, 2), (2, 4) and (3, 8): each step of 1 to the right doubles the height. To the left each step halves the height, and the curve gets closer to the x-axis without touching it.

A logarithm asks for the power

$2^5=32$2⁵ = 32 answers the question: what is 2 to the power 5? The same fact can be read the other way, as a question about the power: what power of 2 gives 32? The answer is 5.

That power has a name. It is the logarithm of 32 to base 2, written ${\log }_{2}32=5$log₂ 32 = 5. In general, ${\log }_{a}x$logₐ x is the power to which a must be raised to give x. So ${\log }_{a}x=y$logₐ x = y means exactly the same as $a^y=x$aʸ = x.

Every logarithm fact is an exponent fact read backward. ${\log }_{2}8=3$log₂ 8 = 3, because $2^3=8$2³ = 8. ${\log }_{3}81=4$log₃ 81 = 4, because $3^4=81$3⁴ = 81. ${\log }_{10}1000=3$log₁₀ 1000 = 3, because $10^3=1000$10³ = 1000.

The answer does not have to be a positive whole number. ${\log }_2\left(\frac{1}{8}\right)=-3$log₂ (1/8) = −3, because $2^{-3}=\frac{1}{8}$2⁻³ = 1/8. ${\log }_{9}3=\frac{1}{2}$log₉ 3 = 1/2, because $9^{\frac{1}{2}}=\sqrt{9}=3$9^½ = √9 = 3.

Raise 2 to the power 5 and the result is 32. Take ${\log }_2$log₂ of 32 and the result is 5, the number the chain started from.

Each undoes the other

Raising 2 to a power and taking ${\log }_2$log₂ are inverse functions: each one undoes the other. Start with 3: $2^3=8$2³ = 8, and ${\log }_{2}8=3$log₂ 8 = 3, which is back where you began. In symbols, ${\log }_2\left(2^x\right)=x$log₂ (2ˣ) = x for every x.

The other order works too. Start with 16: ${\log }_{2}16=4$log₂ 16 = 4, and $2^4=16$2⁴ = 16. So $2^{{\log }_{2}x}=x$2^(log₂ x) = x, for every positive x.

The logarithm of 1

Any number other than 0, raised to the power 0, gives 1. So the power of a that gives 1 is always 0, and ${\log }_{a}1=0$logₐ 1 = 0 in every base: ${\log }_{2}1=0$log₂ 1 = 0, ${\log }_{10}1=0$log₁₀ 1 = 0 and ${\log }_{7}1=0$log₇ 1 = 0.

In the same way $a^1=a$a¹ = a, so ${\log }_{a}a=1$logₐ a = 1 in every base: ${\log }_{2}2=1$log₂ 2 = 1 and ${\log }_{10}10=1$log₁₀ 10 = 1.

No logarithm of 0 or of a negative number

The base of a logarithm is a positive number other than 1. (Every power of 1 is 1, so a base of 1 could only ever give 1.) Every power of a positive base is positive: on the graph of $y=2^x$y = 2ˣ, every point is above the x-axis.

So no power of 2 is 0, and no power of 2 is −4. The questions "what power of 2 gives 0?" and "what power of 2 gives −4?" have no answer, and ${\log }_{2}0$log₂ 0 and ${\log }_2\left(-4\right)$log₂ (−4) do not exist. A logarithm exists only for a positive number.

A positive number less than 1 does have a logarithm. It is negative: ${\log }_2\left(\frac{1}{4}\right)=-2$log₂ (1/4) = −2, because $2^{-2}=\frac{1}{4}$2⁻² = 1/4.

The graph of $y={\log }_{2}x$y = log₂ x

Each point (p, q) on $y=2^x$y = 2ˣ says $2^p=q$2ᵖ = q, which is the same fact as ${\log }_{2}q=p$log₂ q = p. So the point (q, p), with its coordinates swapped, lies on $y={\log }_{2}x$y = log₂ x. (3, 8) on $y=2^x$y = 2ˣ becomes (8, 3) on $y={\log }_{2}x$y = log₂ x, and (0, 1) becomes (1, 0).

Swapping the coordinates of every point reflects a graph in the line y = x. So the graph of $y={\log }_{2}x$y = log₂ x is the reflection of the graph of $y=2^x$y = 2ˣ in the line y = x.

The domain and the range change places. $y=2^x$y = 2ˣ takes any number x and gives only positive values. $y={\log }_{2}x$y = log₂ x takes only positive numbers x, and it can give any value at all. Its graph passes through (1, 0), runs down alongside the y-axis without touching it, and climbs more and more slowly to the right: to go up by 1, x has to double.

The gold curve is $y={\log }_{2}x$y = log₂ x, the white curve is $y=2^x$y = 2ˣ, and the white straight line is y = x. The point (3, 8) on $y=2^x$y = 2ˣ reflects in y = x to (8, 3) on $y={\log }_{2}x$y = log₂ x, and (0, 1) reflects to (1, 0).

Logarithms of other numbers

Most numbers are not an exact power of the base. 20 is not a power of 2, but $2^4=16$2⁴ = 16 and $2^5=32$2⁵ = 32, so ${\log }_{2}20$log₂ 20 lies between 4 and 5. It is about 4.32.

The logarithm to base 10 is called the common logarithm, and it is often written log with no base: log 1000 = 3. A calculator gives it for any positive number. For example, log 2 = 0.3010 to 4 decimal places, which means $10^{0.3010}$10^0.3010 is 2, to that accuracy.

The logarithm of $2\times 10^8$2 × 10⁸

The next problem needs ${\log }_{10}\left(2\times 10^8\right)$log₁₀ (2 × 10⁸). Write each factor as a power of 10. From the calculator value above, $2=10^{0.3010}$2 = 10^0.3010, so $2\times 10^8=10^{0.3010}\times 10^8$2 × 10⁸ = 10^0.3010 × 10⁸. Multiplying powers of the same base adds the indices, so this is $10^{8.3010}$10^8.3010, and ${\log }_{10}\left(2\times 10^8\right)=8.3010$log₁₀ (2 × 10⁸) = 8.3010.

The two logarithms were added: ${\log }_{10}2+{\log }_{10}10^8=0.3010+8=8.3010$log₁₀ 2 + log₁₀ 10⁸ = 0.3010 + 8 = 8.3010. This is a law that holds for every product: the logarithm of a product is the sum of the logarithms.

The usual mistakes

Giving the number instead of the power. ${\log }_{2}32$log₂ 32 is not 32. It is the power of 2 that makes 32, which is 5.

Multiplying or dividing. ${\log }_{2}8$log₂ 8 is not 8 ÷ 2 = 4. It is 3, because 2 × 2 × 2 = 8. In the other direction, $2^5$2⁵ is 32, not 2 × 5 = 10.

Taking ${\log }_{2}1$log₂ 1 to be 1. $2^1=2$2¹ = 2, not 1. The power of 2 that gives 1 is 0, so ${\log }_{2}1=0$log₂ 1 = 0.