Get the same base on both sides and the exponents must match.
Solve for x
2x+1 = 32
Get both sides onto base 2: 32 = 2⁵, so x + 1 = 5 and x = 4. D forgets to undo the +1 — matching the exponents gives you x + 1, which is one more step from the answer.
if bm = bn, then m = n
2x+1 = 32
32 is a power of 2
2x+1 = 25
32 = 2⁵
x + 1 = 5
drop the bases
x = 4
done
2: 2, 4, 8, 16, 32, 64, 128 · 3: 3, 9, 27, 81, 243 · 5: 5, 25, 125, 625. Most exam questions hide inside these.
9 and 27 are both powers of 3.
9x = 27x−1
different bases
(3²)x = (3³)x−1
rewrite both
32x = 33x−3
multiply the exponents
2x = 3x − 3
same base, so equate
x = 3
rearrange
(3³)x−1 is 33(x−1) = 33x−3. And the bracket matters — 3x − 1 would be a different equation entirely.
| Written as | Really is |
|---|---|
| 18 | 2−3 |
| √2 | 21/2 |
| 181 | 3−4 |
| 3√5 | 51/3 |
| (19)x | (3−2)x = 3−2x — keep the minus |
| 1 | b0, for any base |
4x = 18
both are powers of 2
22x = 2−3
rewrite
x = −1.5
2x = −3
Solve (a) 3x + 3x+2 = 90 and (b) 9x − 4 · 3x + 3 = 0.
(a) 3x+2 = 3x · 3² = 9 · 3x
split off the number part
3x(1 + 9) = 90 → 3x = 9
factor out 3x
x = 2
check: 9 + 81 = 90 ✓
(b) 9x = (3x)², so let u = 3x
a quadratic in disguise
u² − 4u + 3 = (u − 1)(u − 3) = 0
u = 1 or u = 3
3x = 1 or 3x = 3 → x = 0 or 1
turn u back into x
{1, 3} are the values of u. And 3x is always positive, so if a factor ever gives u = −4, throw that one away — no x makes 3x negative. Never “add the exponents” in (a): 3x + 3x+2 is not 32x+2.
If 2a = 5, find 8a − 1.
8a − 1 = 8a ÷ 8
split the exponent
8a = (2³)a = (2a)³ = 5³ = 125
rewrite in the given base
8a − 1 = 1258
no logs needed
The moves: a sum in the exponent becomes a product (bm+n = bm · bn), a difference becomes a quotient, and a bigger base becomes a power of the given one.
Take the log of both sides and the exponent comes down.
5x = 20
20 is not a power of 5
log 5x = log 20
log both sides
x log 5 = log 20
the power law
x = log 20 ÷ log 5 ≈ 1.861
5¹ = 5, 5² = 25 — fits
Bringing the exponent down as a coefficient is the entire reason logarithms were invented.
The right side is 1. Solve 5x² − x − 6 = 1.
5x² − x − 6 = 50
1 is any base to the power 0
x² − x − 6 = 0 → (x − 3)(x + 2) = 0
equate exponents, then factor
x = 3 or −2
check: 9 − 3 − 6 = 0 and 4 + 2 − 6 = 0 ✓
What is different: there is no visible power of 5 on the right, and the exponent is a quadratic, so two answers.
A root times a number. Solve 4x = 2√2.
2√2 = 21 · 21/2 = 23/2
multiplying adds the exponents
22x = 23/2
4 = 2²
2x = 32 → x = 34
a fraction is a perfectly good answer
What is different: the right side mixes a whole power and a root; turn both into powers of 2 and add.
One u-value has to go. Solve 4x − 2x − 12 = 0.
u² − u − 12 = 0, with u = 2x
4x = (2x)²
(u − 4)(u + 3) = 0 → u = 4 or u = −3
two values of u
2x = −3 is impossible
a power of 2 is always positive
2x = 4 → x = 2 only
check: 16 − 4 − 12 = 0 ✓
What is different: the quadratic has two roots but the equation has one solution. An equation like 3x = −9 has none at all.
A = A₀ (1 + r)t
multiplier above 1
bacteria, compound interest
multiplier below 1
depreciation, half-life
a colony doubles every hour, starts at 500
multiplier 2
A = 500(2)t
t in hours
after 6 hours: 500(64) = 32,000
2⁶ = 64
“When does it reach 100,000?” turns it back into an exponential equation — and with no common base, that means logs.
Almost every item reduces to a common base of 2, 3 or 5 — fractions bring a minus sign, sums of powers get factored, and 4x or 9x beside 2x or 3x signals u-substitution. The 10-second check: put the answer back into the original equation, not into your rewritten one.
Solve for x
32x−1 = 81
81 = 3⁴, so 2x − 1 = 4, giving 2x = 5 and x = 52. A stops at the exponent instead of solving the little equation inside it. A fractional answer is perfectly normal here.
