Topic · Algebra

Exponential and Logarithmic Functions

When a quantity changes by a fixed percentage instead of a fixed amount, the math becomes exponential. Logarithms are the mirror image — the question "what exponent gets me there?" Together they describe compound interest, radioactive decay, sound intensity, and every system whose rate of change is proportional to its size.

What you'll leave with

  • What separates an exponential function $f(x) = a \cdot b^x$ from a polynomial — and why the variable's position matters.
  • Why the natural base $e \approx 2.71828$ is "natural" and where it comes from.
  • Logarithms as the inverse of exponentials, the three log laws, and change of base.
  • A reusable recipe for solving any exponential or logarithmic equation.
  • How the same machinery describes compound interest, half-life, and pH.

1. What an exponential function is

Exponential function

A function of the form $f(x) = a \cdot b^x$, where the base $b > 0$ and $b \neq 1$, and the coefficient $a \neq 0$. The variable lives in the exponent, not the base.

The whole topic hinges on that last sentence. In $x^2$, the variable is the base and the exponent is fixed — that's a polynomial. In $2^x$, the variable is the exponent and the base is fixed — that's exponential. They look superficially similar; they behave nothing alike.

Polynomial vs exponential

Exponential
  • $2^x$ — base 2, variable in exponent
  • $f(x) = 100 \cdot (1.05)^x$ — 5% growth
  • $y = e^{-x}$ — natural decay
  • $g(t) = 3 \cdot 2^{t/3}$ — doubles every 3
Not exponential
  • $x^2$ — polynomial (variable in base)
  • $x^x$ — not standard form
  • $f(x) = 1^x = 1$ — constant, excluded
  • $(-2)^x$ — base negative, not allowed over $\mathbb{R}$

The growth rule of $f(x) = b^x$ is the defining trait: every time $x$ increases by 1, $f$ is multiplied by $b$. That's a constant ratio per step — not a constant difference, which is what a linear function gives you. The whole shape of the function follows from that one rule.

Laws of exponents (reminder)

These hold for any allowed base $b$ and any real exponents $x, y$. They are not new — they're the same laws you used for integer exponents, extended.

$$ b^{x+y} = b^x \cdot b^y, \qquad b^{x-y} = \frac{b^x}{b^y}, \qquad (b^x)^y = b^{xy}, \qquad b^0 = 1, \qquad b^{-x} = \frac{1}{b^x}. $$

2. Growth vs decay

The base $b$ alone tells you the qualitative behavior:

  • $b > 1$ — exponential growth. The curve climbs ever faster as $x$ increases; for $x \to -\infty$ it flattens onto the x-axis.
  • $0 < b < 1$ — exponential decay. The curve falls ever more slowly toward zero as $x \to \infty$; for $x \to -\infty$ it shoots up.
  • $b = 1$ — excluded. $1^x = 1$ for every $x$, a horizontal line. Not interesting, and breaks the inverse.

Every exponential, regardless of base, shares structure:

  • Domain: all real numbers.
  • Range: $(0, \infty)$ — the function is always positive.
  • Horizontal asymptote at $y = 0$ on one side ($x \to -\infty$ for growth, $x \to +\infty$ for decay).
  • Passes through $(0, a)$ — at $x = 0$, $b^0 = 1$, so $f(0) = a$.
Growth factor, not growth rate

If a quantity grows by 6% per year, the multiplier per year is $1 + 0.06 = 1.06$, not $0.06$. The model is $P(t) = P_0 (1.06)^t$. Confusing factor with rate is the single most common modeling mistake.

Doubling time and half-life

Because the function multiplies by a fixed ratio each step, certain landmarks repeat. For growth, the doubling time is how long it takes for the quantity to double; for decay, the half-life is how long to halve. Both are properties of the rate alone — they don't depend on the starting amount.

If $P(t) = P_0 e^{kt}$ with $k > 0$, then doubling requires $e^{kt_d} = 2$, so $t_d = \dfrac{\ln 2}{k}$. The mirror identity for half-life with $k < 0$ is $t_{1/2} = \dfrac{\ln 2}{|k|}$.

3. The natural base $e$

Of all the possible bases, one stands apart. The number

$$ e = \lim_{n \to \infty}\!\left(1 + \frac{1}{n}\right)^{\!n} \approx 2.71828\ldots $$

is the natural base of exponentials. It first appeared in finance — Jacob Bernoulli (1683) was studying compound interest and noticed that as the compounding frequency goes to infinity, the limit was an irrational number that wouldn't go away. Later Euler (1736) named it, studied it systematically, and proved it irrational.

Why "natural"?

The function $e^x$ has a property no other exponential does: it is its own rate of change. At every point on the curve, the slope equals the height. Any other base $a^x$ has slope $(\ln a) \cdot a^x$ — itself, but with an awkward multiplier in front. Only $e^x$ has that multiplier equal to $1$.

That's why every "rate is proportional to current size" problem — populations, decay, cooling, continuous interest, capacitor discharge — collapses onto $e^{kt}$ once you do the calculus. The $e$ isn't a stylistic choice; it falls out of the equation.

Three equivalent definitions of $e$

You'll see $e$ defined three ways, all of which give the same number. (i) The limit $\lim_{n \to \infty}(1 + 1/n)^n$ — financial origin. (ii) The infinite sum $\sum_{n=0}^{\infty} \tfrac{1}{n!} = 1 + 1 + \tfrac{1}{2} + \tfrac{1}{6} + \cdots$ — Taylor series. (iii) The unique number with $\int_1^e \tfrac{1}{t}\,dt = 1$ — area under $1/t$. Pick whichever your context prefers; calculus stitches them together.

Continuous compounding

Bernoulli's original setup is still the cleanest motivation. Suppose you invest $P$ at annual rate $r$, compounded $n$ times per year for $t$ years:

$$ A = P\!\left(1 + \frac{r}{n}\right)^{nt}. $$

Crank $n$ up — quarterly, monthly, daily, every second — and the limit is:

$$ \lim_{n \to \infty} P\!\left(1 + \frac{r}{n}\right)^{nt} = P e^{rt}. $$

That's continuous compounding. \$1000 at 6% for 5 years grows to $1000 \cdot e^{0.3} \approx \$1349.86$ — only about three dollars more than daily compounding gives you. The cap is $e^{rt}$.

4. Logarithm: the inverse

Exponentials ask "if I start at 1 and multiply by $b$ $x$ times, where do I end up?" Logarithms ask the reverse: "I ended up at $y$ — how many times did I multiply by $b$?" That's all a logarithm is.

Logarithm

$\log_b x = y$ means exactly $b^y = x$. The base $b > 0$ with $b \neq 1$; the argument $x > 0$.

Two pieces of notation worth memorizing once and never mixing up:

  • Natural log: $\ln x = \log_e x$. Used everywhere in calculus, physics, and statistics.
  • Common log: $\log x = \log_{10} x$ in engineering and chemistry contexts. In pure math and many physics texts, $\log$ alone often means $\ln$. Always check the field's convention.
  • Binary log: $\log_2 x$. Native to computer science and information theory.

Two identities that do all the heavy lifting

Because $\log_b$ and $b^{(\cdot)}$ are inverses, applying one undoes the other:

$$ \log_b(b^x) = x \quad\text{for every real } x, \qquad b^{\log_b x} = x \quad\text{for every } x > 0. $$

You will reach for these constantly when solving equations. They're how you "peel" a variable out of an exponent or out from inside a log.

Reading off easy values

ExpressionEqualsBecause
$\log_2 8$$3$$2^3 = 8$
$\log_{10} 1000$$3$$10^3 = 1000$
$\ln e^5$$5$inverse identity
$\log_b 1$$0$$b^0 = 1$ for every $b$
$\log_b b$$1$$b^1 = b$

5. The log laws

Three identities convert products, quotients, and powers inside a log into sums, differences, and coefficients outside. They're the same three laws of exponents in disguise — what's multiplicative on one side is additive on the other.

LawIdentityComes from
Product$\log_b(xy) = \log_b x + \log_b y$$b^{p+q} = b^p \cdot b^q$
Quotient$\log_b\!\left(\dfrac{x}{y}\right) = \log_b x - \log_b y$$b^{p-q} = b^p / b^q$
Power$\log_b(x^n) = n \log_b x$$(b^p)^n = b^{np}$

Two everyday uses:

  • Expand: turn $\log\!\left(\frac{x^2 y}{z^3}\right)$ into $2\log x + \log y - 3 \log z$. Useful when you want to differentiate or simplify.
  • Combine: turn $2\log x + 3\log y$ into $\log(x^2 y^3)$. Useful when solving an equation — collapse everything into a single log first.
There is no law for $\log(x + y)$

$\log(x + y)$ does not equal $\log x + \log y$. The product law fires for $\log(xy)$, not $\log(x+y)$. This is the single most common error in early use of logs — and it's why the laws are stated for products, not sums.

Why logs were invented

Before electronic calculators, multiplying two seven-digit numbers was a half-hour ordeal. Napier (1614) noticed that if you tabulate $\log x$ for many $x$, then "multiply $a \cdot b$" becomes "look up $\log a$, look up $\log b$, add them, look up the antilog." Addition is easy. Multiplication is hard. The log table — and later the slide rule — used this identity to dominate engineering computation for 350 years.

6. Change of base

Your calculator has $\ln$ and $\log_{10}$. What if you need $\log_7 200$? The change-of-base formula bridges any two bases:

$$ \log_b x = \frac{\log_c x}{\log_c b} \qquad \text{for any valid base } c. $$

In practice this means: pick whichever base your calculator supports — usually $e$ or $10$ — and divide.

$$ \log_7 200 = \frac{\ln 200}{\ln 7} \approx \frac{5.298}{1.946} \approx 2.723. $$

The denominator carries the base. Mixing up which goes on top and which on the bottom is a frequent slip — remember that $\log_b b = 1$, so the base appears in the denominator (where it equals $\log_c b$).

Why it works

Let $y = \log_b x$, so $b^y = x$. Take $\log_c$ of both sides: $\log_c(b^y) = \log_c x$, then by the power law $y \log_c b = \log_c x$, so $y = \log_c x / \log_c b$. One application of the power law is the whole proof.

7. Playground: $f(x) = a \cdot b^x$

Drag the base $b$ between $0.3$ and $4$. Watch growth become decay as $b$ crosses $1$. Adjust $a$ to scale the curve vertically. The horizontal line at $y = 0$ is the asymptote — the curve never crosses it.

f(x) = 1.0 · 2.00x
growth  ·  f(1) = 2.00  ·  f(0) = 1.00
1.0
2.00
x y −4 −2 2 4 6 4 2 −2
Try it

Set $a = 1$ and slide $b$ from $4$ down through $1$ to $0.3$. Watch the curve flatten as $b \to 1$, then flip its direction. The transition point $b = 1$ is the single horizontal line $y = 1$ — that's why $b = 1$ is excluded: there's no exponential behavior to speak of.

8. Solving exponential & log equations

Two patterns cover almost everything. Pick the one that matches your equation, then run the recipe.

If both sides can be written as the same base raised to something, exponentials are injective: the exponents must be equal.

  1. Rewrite both sides as powers of a common base.
  2. Set the exponents equal.
  3. Solve for the variable.

Example: $2^{x+1} = 8$. Rewrite $8$ as $2^3$, so $2^{x+1} = 2^3$, giving $x + 1 = 3$, hence $x = 2$.

When you can't match bases, apply a logarithm to both sides. The power law brings the exponent down where you can solve for it.

  1. Isolate the exponential term.
  2. Take $\ln$ (or any $\log_b$) of both sides.
  3. Use the power law to pull the exponent out front.
  4. Solve the resulting linear equation.

Example: $5^x = 20$. Take $\ln$: $x \ln 5 = \ln 20$, so $x = \ln 20 / \ln 5 \approx 1.861$.

When the variable is inside a logarithm, work in the opposite direction: collapse to one log, then exponentiate.

  1. Use log laws to combine into a single $\log$ term.
  2. Exponentiate both sides (raise $b$ to both sides) to remove the log.
  3. Solve the resulting polynomial.
  4. Check each candidate: the original log's argument must be positive.

Example: $\log x + \log(x - 3) = 1$ (base 10). Combine: $\log[x(x-3)] = 1$, so $x(x-3) = 10$, i.e. $x^2 - 3x - 10 = 0$, giving $(x-5)(x+2)=0$. Reject $x = -2$ (would make $\log(-2)$ undefined). Answer: $x = 5$.

Always check for extraneous solutions

Exponentiating a log equation can introduce solutions that don't satisfy the original. The domain of $\log_b x$ is $x > 0$; any candidate that puts a non-positive number inside a log is invalid and must be discarded.

9. Applications

The same function — $f(t) = a \cdot b^t$ or its continuous cousin $a e^{kt}$ — describes problems that look unrelated at first.

Compound interest

Principal $P$ at annual rate $r$, compounded $n$ times per year for $t$ years:

$$ A = P\!\left(1 + \tfrac{r}{n}\right)^{nt}. $$

Annual: $n = 1$. Quarterly: $n = 4$. Monthly: $n = 12$. Daily: $n = 365$. Continuous: $n \to \infty$, giving $A = Pe^{rt}$.

Radioactive decay and half-life

For a sample of $N_0$ atoms decaying with half-life $t_{1/2}$:

$$ N(t) = N_0 \cdot \left(\tfrac{1}{2}\right)^{t/t_{1/2}} = N_0 \, e^{-kt}, \qquad k = \tfrac{\ln 2}{t_{1/2}}. $$

Carbon-14 has $t_{1/2} \approx 5730$ years. A sample with 25% of its original C-14 has gone through two half-lives, so it's about $11{,}460$ years old.

pH and other logarithmic scales

The pH of a solution is

$$ \text{pH} = -\log_{10}[\text{H}^+], $$

where $[\text{H}^+]$ is the hydrogen-ion concentration in moles per liter. A drop in pH by $1$ means $[\text{H}^+]$ is $10\times$ higher. The same idea drives decibels (sound intensity) and the Richter scale (earthquake amplitude): when a quantity spans many orders of magnitude, taking a log gives you a scale humans can reason with.

Why log scales

Sound intensities range over $10^{12}$ from a whisper to a jet engine; on a linear scale, every conversation pegs the bottom of the chart. Decibels — $10 \log_{10}(I/I_0)$ — compress that into 0 to 120. Whenever your data has multiplicative structure, a log scale makes the structure visible.

10. Common pitfalls

$\log(x + y) \neq \log x + \log y$

The product law is for $\log(xy)$. There is no analogous identity for $\log$ of a sum — leave it alone.

$\log(x^n) \neq (\log x)^n$

The power law moves the exponent out front: $\log(x^n) = n \log x$. The exponent does not apply to the whole log expression.

Growth rate is not growth factor

"5% annual growth" means a multiplier of $1.05$ per year, not $0.05$. The model is $P_0(1.05)^t$, never $P_0(0.05)^t$. Same trap for decay: 5% annual decay is multiplier $0.95$, not $-0.05$.

Wrong base in change of base

$\log_b x = \dfrac{\ln x}{\ln b}$, not $\dfrac{\ln b}{\ln x}$. Sanity check: $\log_b b$ must equal $1$, which happens only when the base lives in the denominator.

Forgetting domain restrictions

$\log_b x$ requires $x > 0$. After solving any log equation, plug each candidate back into the original equation and discard those that make a log argument zero or negative.

$\log$ vs $\ln$ conventions clash

In engineering and chemistry, $\log$ usually means $\log_{10}$. In pure math and many physics texts, $\log$ alone often means $\ln$. When in doubt, write $\log_{10}$ or $\ln$ explicitly — a single ambiguous symbol can flip an answer by a factor of $\ln 10 \approx 2.303$.

11. Worked examples

Work each one before opening the solution. Reading the answer is not learning — the recipe lives in your fingers, not the page.

Example 1 · Solve $3^{2x - 1} = 81$

Step 1. Rewrite $81$ in base $3$: $81 = 3^4$, so

$$ 3^{2x - 1} = 3^4. $$

Step 2. Match exponents:

$$ 2x - 1 = 4 \quad\Longrightarrow\quad x = \tfrac{5}{2}. $$

Check. $3^{2(5/2) - 1} = 3^4 = 81$ ✓

Example 2 · Solve $5 e^{0.2x} = 40$

Step 1. Isolate the exponential:

$$ e^{0.2x} = 8. $$

Step 2. Take $\ln$ of both sides:

$$ 0.2x = \ln 8. $$

Step 3. Solve:

$$ x = \frac{\ln 8}{0.2} \approx \frac{2.079}{0.2} \approx 10.40. $$
Example 3 · Combine $2\log_2 x - \log_2 (x+1)$ into one log

Step 1. Pull the coefficient inside via the power law:

$$ 2\log_2 x = \log_2(x^2). $$

Step 2. Apply the quotient law:

$$ \log_2(x^2) - \log_2(x+1) = \log_2\!\left(\frac{x^2}{x + 1}\right). $$
Example 4 · How long until \$1000 doubles at 5% continuous?

Setup. Continuous compounding: $A = P e^{rt}$. Doubling means $A = 2P$, so $e^{0.05 t} = 2$.

Solve. Take $\ln$:

$$ 0.05 t = \ln 2 \quad\Longrightarrow\quad t = \frac{\ln 2}{0.05} \approx \frac{0.693}{0.05} \approx 13.86 \text{ years}. $$

This is the source of the so-called "rule of 72": $72 / r\%$ approximates the doubling time because $\ln 2 \approx 0.693$ and $0.693 \times 100 / r \approx 69/r$, rounded up to $72$ for friendlier divisors.

Example 5 · Solve $\log_3(x) + \log_3(x - 6) = 3$

Step 1. Combine via the product law:

$$ \log_3[x(x - 6)] = 3. $$

Step 2. Exponentiate (apply $3^{(\cdot)}$) to remove the log:

$$ x(x - 6) = 3^3 = 27. $$

Step 3. Solve the quadratic:

$$ x^2 - 6x - 27 = 0 \quad\Longrightarrow\quad (x - 9)(x + 3) = 0. $$

Step 4. Check domain. $x = -3$ makes $\log_3(-3)$ undefined — reject. $x = 9$ gives $\log_3 9 + \log_3 3 = 2 + 1 = 3$ ✓. Answer: $x = 9$.

Example 6 · A sample has 12.5% of its original carbon-14. How old?

Setup. $N(t)/N_0 = (1/2)^{t/5730} = 0.125 = 1/8$.

Solve. $1/8 = (1/2)^3$, so $t/5730 = 3$, giving

$$ t = 3 \times 5730 = 17{,}190 \text{ years}. $$

Three half-lives have elapsed.

Sources & further reading

The content above is synthesized from established references. When a step here looks ambiguous, treat the primary sources as ground truth — they're where the formal statements live.

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