1
In an exponential function the variable is the exponent.
f(x) = bx with b > 0 and b ≠ 1.
f(x) = 2x
exponential ✓
x is upstairs
f(x) = x²
quadratic ✗
x is the base
f(x) = 1x
not allowed ✗
it is just the line y = 1
Base 1 is barred because 1 to any power is 1, which is a horizontal line and has no
inverse. Negative bases are barred because
(−4)1/2 has no real value.
2
Same base, equal exponents.
32x = 81
81 is a power of 3
32x = 3⁴
rewrite the right side
2x = 4
bases match, so exponents do
This only works once the bases are identical
From 2x = 3x you may not conclude x = x and stop; the only solution is x = 0.
Matching exponents is a licence you earn by matching bases first.
3
Rewrite both sides onto a common base.
4x = 8
neither is a power of the other
(2²)x = 2³
both are powers of 2
22x = 2³
multiply the exponents
2x = 3, so x = 32
a fraction is a fine answer
Look for the smallest base both numbers share. For 4 and 8 that is 2; for 9 and 27
it is 3; for 25 and 125 it is 5.
4
An inequality keeps its direction only if the base is above 1.
x > 5
direction unchanged
(12)x > (12)⁵
base 12, shrinking
Test it: (12)³ = 0.125 and (12)⁵ = 0.031. The smaller exponent gave the bigger
value, because each extra power halves the number.
5
Variations: a fraction on the right, a 1 on the right, x on both sides.
A fraction on the right: 2x = 18
18 = 12³ = 2−3
one over a power is a negative power
x = −3
a negative answer is fine
Check: 2−3 = 12³ = 18 ✓. A fraction below 1 with a base above 1 always needs a negative exponent.
A 1 on the right: 7x − 2 = 1
1 = 7⁰
any non-zero base to the power 0 is 1
x − 2 = 0
equate the exponents
1 is a power of every base. Answering "no solution" because 1 is not a multiple of 7 is the slip.
x on both sides: 4x + 1 = 8x
22(x + 1) = 23x
both sides onto base 2
2x + 2 = 3x
multiply out the left exponent
Check: 4³ = 64 and 8² = 64 ✓. The bracket matters: 2(x + 1) is 2x + 2, not 2x + 1.
6
Growth and decay are the reason this topic exists.
V(t) = 20000(0.75)t
keeps 75% each year
V(1) = 15000, V(2) = 11250
still above 10,000
V(3) = 8437.50
below, so in the 3rd year
x = 4 for the opening problem
Losing 25% a year is not losing all of it in four years. Each year takes a
quarter of a smaller amount, so the value approaches zero without ever arriving —
which is exactly what the horizontal asymptote in the next lesson describes.
On the exam
Periodical exams ask you to tell an exponential function from a power function, solve an equation by matching bases, or solve a short inequality; entrance tests favour bases that are not powers of each other (8 and 4, 27 and 9) and the base-below-1 inequality. Rewrite both sides on the smallest shared base, then equate the exponents. For a 10-second check, put your x back into the original and see that both sides give the same number.
Remember the flow
Variable in the exponent · Match the bases · Equate the exponents · Flip the inequality if the base is below 1
If the bases will not match, logarithms are the tool — that is two lessons away.
