Two functions, one mirror. An exponential races upward by repeated multiplication; a logarithm walks it backward and asks a single quiet question — "to what power?" Learn to see them as reflections of each other and the whole topic falls into place.
Exponential growth: repeated multiplication
An exponential function has the form \( y = b^{x} \), where the fixed number \( b \) is the base and the variable \( x \) sits up in the exponent. The exponent counts how many times you multiply the base by itself. With base \( 2 \), each step doubles the value:
\[ 2^{0}=1,\quad 2^{1}=2,\quad 2^{2}=4,\quad 2^{3}=8,\quad 2^{4}=16,\quad 2^{5}=32 \]With base \( 3 \) the value triples at every step: \( 3^{0}=1,\ 3^{1}=3,\ 3^{2}=9,\ 3^{3}=27 \). Notice what makes exponential growth so dramatic — you are not adding a fixed amount each time, you are multiplying by the base again and again. That is why the curve starts gently and then turns shockingly steep: every step is bigger than the last by the same factor.
If you want to refresh the plain idea of an exponent — a base multiplied by itself a whole number of times — drag the exponent here and watch \( 2^{x} \) build up from \( 1 \):
Adding versus multiplying. A straight line grows by adding the same amount each step. An exponential grows by multiplying by the same factor each step. That single difference is why \( 2^{x} \) overtakes any straight line eventually — and why it does so faster than almost anyone expects.
The big idea: a logarithm runs it backwards
Here is the secret that makes logarithms easy. A logarithm is simply an exponent wearing a question mark. Where the exponential asks "I have the power \( x \) — what value do I get?", the logarithm asks the reverse:
\[ \log_{b}(x)\ =\ \text{"} b \text{ to } \mathbf{what}\text{ power gives } x \text{?"} \]So \( \log_{2}(8) \) means: \( 2 \) raised to what power gives \( 8 \)? Since \( 2^{3}=8 \), the answer is \( 3 \). We write \( \log_{2}(8)=3 \). That is the whole definition. A log is just an exponent — the exponent you were hunting for.
The two statements below say exactly the same thing, read in opposite directions:
\[ b^{y}=x \qquad\Longleftrightarrow\qquad \log_{b}(x)=y \]Read left to right, you start with the power and find the value. Read right to left, you start with the value and recover the power. Each undoes the other — they are inverse functions.
Tip Whenever a logarithm confuses you, translate it back into the exponential question. "\( \log_{5}(25)=\,? \)" becomes "\( 5 \) to what power is \( 25 \)?" The answer, \( 2 \), is now obvious. The exponential form is your safety net.
Seeing the mirror: two curves, one line
Because the two functions undo each other, their graphs are reflections across the dashed line \( y = x \). Whatever \( y = b^{x} \) does going up, \( y = \log_{b} x \) does the same thing turned sideways. Toggle between exp, log, and both below, and try changing the base — watch the two curves stay locked together as perfect mirror images:
Look for the anchor points. Every exponential \( y=b^{x} \) passes through \( (0,1) \), because \( b^{0}=1 \) for any base. Reflect that point across \( y=x \) and you get \( (1,0) \) — which is exactly where every logarithm \( y=\log_{b} x \) crosses the axis, because \( \log_{b}(1)=0 \). The pair \( (0,1)\leftrightarrow(1,0) \) is the mirror caught in the act. Likewise \( (1,b) \) on the exponential reflects to \( (b,1) \) on the log.
Two facts to memorise
These two come straight out of the definition, and they cover a surprising number of problems on their own:
\[ \log_{b}(1)=0 \qquad\text{because}\qquad b^{0}=1 \] \[ \log_{b}(b)=1 \qquad\text{because}\qquad b^{1}=b \]In words: the log of \( 1 \) is always \( 0 \) (you need no copies of the base to make \( 1 \)), and the log of the base itself is always \( 1 \) (you need exactly one copy). True for base \( 2 \), base \( 10 \), base \( 7 \) — every base. No memorising of special cases required, just the question "what power?"
The three log laws: turning × into +
Logarithms have a near-magical property: they convert multiplication into addition. Here are the three laws, the engine of everything you will do with logs:
\[ \log_{b}(xy) = \log_{b} x + \log_{b} y \] \[ \log_{b}\!\left(\tfrac{x}{y}\right) = \log_{b} x - \log_{b} y \] \[ \log_{b}(x^{n}) = n\,\log_{b} x \]Each one is just an exponent rule in disguise. Multiplying powers adds their exponents (\( b^{m}\cdot b^{n}=b^{m+n} \)), so taking the log of a product adds the logs. Dividing powers subtracts exponents, so dividing inside a log subtracts. And a power of a power multiplies exponents, so an exponent inside a log comes out front as a multiplier.
This is literally why logarithms were invented. Four centuries ago, John Napier built log tables so that astronomers and navigators could replace slow, error-prone multiplication of huge numbers with quick addition — turning days of arithmetic into hours. The same trick lives on today wherever quantities span enormous ranges: the pH scale of acidity, the decibel scale of loudness, and the Richter scale of earthquakes are all logarithmic, so that each whole step means a tenfold change.
Why one step means ten times. On the Richter scale a magnitude \( 6 \) quake is not a little stronger than a magnitude \( 5 \) — it releases about \( 10 \) times the ground motion, because the scale is \( \log_{10} \). Logarithms let a single, readable number stand in for a colossal range.
Worked examples
- The unknown is stuck up in the exponent, so take a logarithm to bring it down. By definition, \( 2^{x}=32 \) means \( x = \log_{2} 32 \).
- Ask the log's question: \( 2 \) to what power gives \( 32 \)?
- Count the doublings: \( 2,4,8,16,32 \) — that is five of them, so \( 2^{5}=32 \).
Therefore \( x = \log_{2} 32 = \mathbf{5} \). Check: \( 2^{5}=32 \). ✓
- Translate to the exponential question: \( 10 \) to what power gives \( 1000 \)?
- Write \( 1000 \) as a power of \( 10 \): \( 1000 = 10 \times 10 \times 10 = 10^{3} \).
- So the power we need is \( 3 \).
Therefore \( \log_{10} 1000 = \mathbf{3} \). A base-\( 10 \) log just counts the zeros: \( 1000 \) has three. ✓
Practice
Try each one yourself, then reveal the full solution.
1. Evaluate \( \log_{3} 81 \).
Translate to the exponential question: \( 3 \) to what power gives \( 81 \)?
Build up the powers of \( 3 \): \( 3^{1}=3,\ 3^{2}=9,\ 3^{3}=27,\ 3^{4}=81 \).
So \( 3^{4}=81 \), which means \( \log_{3} 81 = \mathbf{4} \).
2. Evaluate \( \log_{5} 1 \).
This is one of the two facts to memorise: the log of \( 1 \) is always \( 0 \).
Why? Because \( 5^{0}=1 \) — you need no copies of the base to make \( 1 \).
So \( \log_{5} 1 = \mathbf{0} \).
3. Solve \( 2^{x} = 16 \).
The unknown is in the exponent, so rewrite as a logarithm: \( x = \log_{2} 16 \).
Ask: \( 2 \) to what power gives \( 16 \)? Count the doublings: \( 2,4,8,16 \) — that is four of them.
Since \( 2^{4}=16 \), we get \( x = \mathbf{4} \).