Exponential & Log Functions

A2.F.1Graph exponentials01/04

Graph exponential functions.

Step01of 04

General form. $f (x) = a \cdot b^{x}$ with $b > 0$ , $b \neq = 1$ .

$b > 1$	→growth — graph rises sharply to the right
$0 < b < 1$	→decay — graph falls to the right
a	→y-intercept (output at $x = 0$ )
Horizontal asymptote	→ $y = 0$ (the function approaches but never touches the x-axis)

A2.F.1 — Graph exponential functions.

Step 01 of 04

General form. $f (x) = a \cdot b^{x}$ with $b > 0$ , $b \neq = 1$ .

$b > 1$	→growth — graph rises sharply to the right
$0 < b < 1$	→decay — graph falls to the right
a	→y-intercept (output at $x = 0$ )
Horizontal asymptote	→ $y = 0$ (the function approaches but never touches the x-axis)

Step 02 of 04

Visualize a growth and a decay curve side by side.

Forest:

y = 2^{x}

. Grey:

y = (0.5)^{x}

. Both pass through (0, 1).

Step 03 of 04

Domain and range.

Domain	→all real numbers
Range	→positive reals $(0, \infty)$
Y-intercept	→ $(0, a)$
X-intercept	→none — exponentials never cross the x-axis

Step 04 of 04

Transformations. $f (x) = a \cdot b^{x - h} + k$ shifts the parent graph right by $h$ , up by $k$ . The horizontal asymptote moves to $y = k$ .

Key insight

Exponential functions: positive base > 1 grows, base between 0 and 1 decays. Y-intercept = a, horizontal asymptote at

y = 0

(or

y = k

after vertical shift).

A2.F.2 — Solve exponential equations.

Step 01 of 04

Strategy 1 — same base. If both sides can be rewritten with the same base, set the exponents equal.

2^{x + 1} = 8 \Rightarrow 2^{x + 1} = 2^{3} \Rightarrow x + 1 = 3 \Rightarrow x = 2

Works whenever the constants on both sides are powers of a common base.

Step 02 of 04

Worked example. Solve $9^{x - 1} = 2 7^{2 x}$ .

Both 9 and 27 are powers of 3:

(3^{2})^{x - 1} = (3^{3})^{2 x}

3^{2 x - 2} = 3^{6 x}

2 x - 2 = 6 x \Rightarrow x = - \frac{1}{2}

Step 03 of 04

Strategy 2 — take a logarithm. When you can't get a common base, apply $lo g$ (base 10) or $ln$ (base $e$ ) to both sides.

5^{x} = 17

lo g (5^{x}) = lo g (17)

x lo g 5 = lo g 17 \Rightarrow x = \frac{lo g 17}{lo g 5} \approx 1.76

The " $lo g A^{B} = B lo g A$ " property is what pulls $x$ down from the exponent.

Step 04 of 04

Mixed example. Solve $3^{2 x} = 5 \cdot 3^{x} - 6$ .

Let $u = 3^{x}$ . Then $u^{2} = 5 u - 6$ :

u^{2} - 5 u + 6 = 0 \Rightarrow u = 2 or u = 3

Back-substitute:

3^{x} = 2 \Rightarrow x = lo g_{3} 2 or 3^{x} = 3 \Rightarrow x = 1

Key insight

Same base → set exponents equal. Different bases → take

lo g

of both sides and use the power property to pull

x

out of the exponent. Equations like

a^{2 x} - 5 a^{x} + 6 = 0

substitute to a quadratic.

A2.F.3 — Definition and graph of logarithms.

Step 01 of 04

A logarithm is just an exponent. $lo g_{b} x$ answers "what power of $b$ equals $x$ ?"

lo g_{b} x = y ⟺ b^{y} = x

Logarithms and exponentials are INVERSE functions — each undoes the other.

$lo g_{2} 8$	→3 (because 2³ = 8)
$lo g_{5} 25$	→2 (because 5² = 25)
$lo g_{10} 1000$	→3
$lo g_{b} 1$	→0 (any positive base, b⁰ = 1)
$lo g_{b} b$	→1

Step 02 of 04

Special bases.

$lo g x$	→"common log" — base 10 understood
$ln x$	→"natural log" — base $e \approx 2.718$

The natural log appears constantly in calculus and growth/decay problems because it pairs with $e^{x}$ .

Step 03 of 04

Graph of $y = lo g_{b} x$ (for $b > 1$ ):

y = lo g_{2} x

— vertical asymptote at

x = 0

, x-intercept at

(1, 0)

Step 04 of 04

Domain, range, intercepts.

Domain	→positive reals $(0, \infty)$ — log of a non-positive is undefined
Range	→all reals
Vertical asymptote	→ $x = 0$
X-intercept	→ $(1, 0)$ — for any base
Y-intercept	→none

Notice the perfect mirror with exponentials: their domain/range swap, their asymptotes swap, their intercepts swap. Inverse functions are reflections across $y = x$ .

Key insight

lo g_{b} x

= "the exponent on

b

that gives

x

." Domain is positive reals only. Vertical asymptote at

x = 0

, x-intercept at

(1, 0)

. Logs and exponentials are inverses — same shape, swapped axes.

A2.F.4 — Properties of logarithms.

Step 01 of 04

Three core properties, all derived from the corresponding exponent properties (and the fact that logs ARE exponents).

Product	→ $lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$
Quotient	→ $lo g_{b} (M / N) = lo g_{b} M - lo g_{b} N$
Power	→ $lo g_{b} (M^{k}) = k lo g_{b} M$

These let you BREAK APART or COMBINE log expressions to simplify or solve.

Step 02 of 04

Expand a log expression.

lo g (\frac{x ^{3} y}{z ^{2}})

= lo g x^{3} + lo g y - lo g z^{2}

= 3 lo g x + \frac{1}{2} lo g y - 2 lo g z

Step 03 of 04

Condense into a single log.

2 lo g x + 3 lo g y - lo g z

= lo g x^{2} + lo g y^{3} - lo g z

= lo g (\frac{x ^{2} y ^{3}}{z})

Step 04 of 04

Change of base. Calculators only do $lo g_{10}$ and $ln$ . To compute a log in a different base, convert:

lo g_{b} x = \frac{lo g x}{lo g b} = \frac{ln x}{ln b}

Example: $lo g_{3} 50 = \frac{lo g 50}{lo g 3} \approx 3.56$ .

Key insight

Three log properties — product (sum), quotient (difference), power (coefficient out front). Use them to expand for derivative-style work or to condense for solving. Change of base lets a calculator handle any base.

A2.F.5 — Solve logarithmic equations.

Step 01 of 04

Strategy 1 — same log on both sides. Set the arguments equal.

lo g_{2} (x + 3) = lo g_{2} (2 x - 1)

x + 3 = 2 x - 1 \Rightarrow x = 4

Always check that the answer keeps the original log arguments POSITIVE. Here, both $4 + 3 = 7$ and $2 (4) - 1 = 7$ are positive, so $x = 4$ is valid.

Step 02 of 04

Strategy 2 — convert to exponential form. When only one side has a log, rewrite using the definition.

lo g_{4} (x + 1) = 3 \Rightarrow 4^{3} = x + 1 \Rightarrow x = 63

Step 03 of 04

Strategy 3 — combine logs first, then solve.

lo g (x) + lo g (x - 3) = 1

lo g (x (x - 3)) = 1 \Rightarrow x (x - 3) = 10

x^{2} - 3 x - 10 = 0 \Rightarrow (x - 5) (x + 2) = 0

x = 5 or x = - 2

Step 04 of 04

Always check. $x = - 2$ would make $lo g (- 2)$ undefined — extraneous. Only $x = 5$ is valid.

The argument-positive trap. Logarithm domains exclude non-positive numbers. ANY answer that makes any original log argument zero or negative is automatically rejected.

Key insight

Combine logs (when possible) → convert to exponential or set arguments equal → solve → reject any answer making a log argument ≤ 0. The extraneous-solution check is non-negotiable.

A2.F.6 — Exponential and log word problems.

Step 01 of 04

Most exp/log word problems boil down to plugging into one of two templates and solving for the unknown.

Continuous growth	→ $A = P e^{r t}$
Periodic compounding	→ $A = P (1 + \frac{r}{n})^{n t}$
Half-life decay	→ $A = P \cdot (\frac{1}{2})^{t / T}$ where $T$ = half-life
pH / decibels / Richter	→logarithmic scales — apply $lo g$ directly

Step 02 of 04

Worked example — half-life. A 200-mg sample of carbon-14 has a half-life of 5,730 years. How much remains after 10,000 years?

A = 200 \cdot (\frac{1}{2})^{10000/5730} \approx 200 (0.297) \approx 59.4 mg

The fractional exponent represents how many half-lives have elapsed.

Step 03 of 04

Worked example — solve for time. $1,000 invested at 5% continuous interest. When does it double?

2000 = 1000 e^{0.05 t}

2 = e^{0.05 t}

ln 2 = 0.05 t \Rightarrow t = \frac{ln 2}{0.05} \approx 13.86 years

The "rule of 72" approximation: $t \approx 72/ r_{%} = 72/5 = 14.4$ years — close.

Step 04 of 04

Worked example — pH. pH = $- lo g [H^{+}]$ where $[H^{+}]$ is hydrogen-ion concentration.

A solution with pH 4 has $[H^{+}] = 1 0^{- 4}$ moles/liter; pH 7 has $1 0^{- 7}$ . Each pH unit = 10× change in concentration.

Key insight

Pick the right model from the problem's wording (continuous, compound, half-life, log scale). Plug knowns in. If the unknown is in the exponent, take a log. If the unknown is the base or coefficient, isolate by algebra. Always check units.

Free diagnostic

Test your understanding: 10 questions, ~10 min.

Pulled live from our bank for A2.F · scored instantly with worked solutions · free, no sign-up.

Start diagnostic →Print worksheets

A2.F.1 – A2.F.6 · CCSS F.IF.C.7e / F.LE.A.4 / F.BF.B.5·Full curriculum →