Combine into one log, then undo it — a log is just an exponent in disguise.
Solve for x
log3(x) + log3(2) = 2
Adding logs multiplies the insides: log₃(2x) = 2, so 2x = 3² = 9 and x = 4.5. D forgets to divide by the 2 after undoing the log.
“What power turns the base into this number?”
log3(9) = 2
32 = 9
the small 3
what sits inside
what the log equals
Whatever is inside a log must be positive. So x > 0 here, and any answer that breaks that gets thrown away at the end.
The same reading solves for an unknown value. log₃₂ 4 = x means 32ˣ = 4. Both are powers of 2: 2⁵ˣ = 2², so 5x = 2 and x = 25.
You cannot undo two logs at once, so squeeze them into one.
log3(x) + log3(2) = 2
log3(x · 2) = 2
which tidies up to
log3(2x) = 2
logb(M) + logb(N) = logb(MN)
It only works when both bases match.
One log alone on the left — now it can be unwrapped.
log3(2x) = 2
base 3, number 2x, power 2
32 = 2x
3² = 9
2x
The log sign is gone, so everything after this is ordinary algebra.
A log inside a log unwraps twice, outside first. log₃ (log₂ x) = 1 gives log₂ x = 3¹ = 3, then x = 2³ = 8.
3² = 2x
from Step 3
9 = 2x
3 · 3 = 9
x = 4.5
÷ 2 both sides
4.5 is positive, so the log of it exists and the answer stands. A negative or zero answer would have to be rejected.
Solve log₂ x + log₂ (x + 2) = 3.
log₂ [x(x + 2)] = 3
one log first
x² + 2x = 2³ = 8
unwrap with the base
(x + 4)(x − 2) = 0 → x = −4 or x = 2
two candidates
x = 2 only
log₂ (−4) does not exist
x = −4 makes x(x + 2) = 8 positive, so the combined log looks fine — but the original log₂ x is undefined. Check: log₂ 2 + log₂ 4 = 1 + 2 = 3 ✓. When the original has a single log of x² − 16, say, both signs of x can survive.
If log M = log N (same base), then M = N. Solve log (2x + 6) − log (x − 1) = log 5.
log [2x + 6x − 1] = log 5
minus → divide
2x + 6x − 1 = 5
equal logs, equal insides
2x + 6 = 5x − 5 → 3x = 11
clear the fraction
x = 113
x − 1 = 83 > 0, so it stands
If the right side is a sum such as log 2 + log 7, condense it to log 14 first. Never “cancel the logs” before each side is a single log: log (2x + 6) − log (x − 1) is not log (2x + 6 − x + 1).
The unknown is the base. Solve logx 49 = 2.
x² = 49
base to the power equals the inside
x = 7 or x = −7
both square to 49
x = 7 only
a base must be positive and not 1
What is different: the rejection comes from the base rule, not from the inside of the log.
The log equals a negative. Solve log₂ (x − 1) = −3.
x − 1 = 2−3 = 18
unwrap as usual
x = 98
inside: 98 − 1 = 18 > 0 ✓
What is different: a negative VALUE of the log is fine. Only a negative or zero INSIDE is banned, so nothing is rejected here.
The domain rule is about the inside, x − 1. The log itself may be any real number: log₂ (18) = −3 is perfectly true.
Every candidate fails. Solve log₂ (x + 1) = log₂ (2x + 5).
x + 1 = 2x + 5
one log each side: drop them
x = −4
plain algebra
x + 1 = −3 < 0 → no solution
the only candidate is rejected
What is different: the algebra gives an answer, but the equation has none. The solution set is empty, ∅.
Every log equation uses one of these to get down to a single log.
| If the logs are | The rule | Inside, that means |
|---|---|---|
| added | logb(M) + logb(N) = logb(MN) | multiply |
| subtracted | logb(M) − logb(N) = logb(MN) | divide |
| in front, a number | k · logb(M) = logb(Mk) | becomes a power |
We used the first one back on Step 2.
All three need the same base on every log. Different bases must be converted first.
Nearly every log equation is “combine into one log, unwrap, solve a linear or quadratic, reject”. The wrong choices are usually the rejected root or a set with both roots, so the 10-second check is plugging each candidate into the original logs and asking: is every inside positive?
log3(4.5) + log3(2) = log3(9)
Solve for x
log2(x) + log2(x − 2) = 3
Combine: log₂(x(x − 2)) = 3, so x² − 2x = 8 and (x − 4)(x + 2) = 0. That offers 4 and −2 — but a logarithm cannot take a negative or zero input, so −2 is rejected. A keeps the invalid root; always test solutions back in the original log.
Hint: x appears in both logs, so combining gives a quadratic. You will get two answers — only one survives the domain check.
