Every answer has to be fed back in. A log of a negative number is not a solution.
Solve for x
log₂ (x − 1) = 3
Rewrite in exponential form: x − 1 = 2³ = 8, so x = 9. Check the domain: x − 1 = 8, which is positive, so it stands. B forgets to add the 1 back and answers the argument rather than x. A multiplies 2 by … nothing that applies here. Confirm: log₂ 8 = 3 ✓
log₅ x = 2
"5 to what power is x?"
x = 5²
the answer is the power, so undo it
x = 25
positive, so it is allowed
The base stays the base and the number on the right becomes the exponent. That single move solves most of these.
log(2x) = log(x + 4)
same base on both sides
2x = x + 4
drop the logs
x = 4
check: 8 and 8, both positive ✓
This works because the logarithm is one-to-one: two inputs can never produce the same log, so equal logs force equal arguments.
log x + log(x − 9) = 1
base 10 assumed
log(x(x − 9)) = 1
product law
x² − 9x = 10
rewrite as a power of 10
x² − 9x − 10 = 0, (x − 10)(x + 1) = 0
two candidates
Two candidates, and only one survives. Which one is the whole point of the next step.
Combining the logs threw away the information that x and x − 9 both had to be positive. The quadratic happily returns −1, which the original equation could never accept. Writing both roots down is the error this whole lesson exists to prevent.
The number on the right is still the power; the only change is that a whole expression must equal that power.
Solve log₃ (2x − 1) = 2.
2x − 1 = 3² = 9
the whole argument equals the power
2x = 10, so x = 5
then solve the linear equation
argument 2(5) − 1 = 9 > 0 ✓
check the domain
Writing 2x − 1 = 3 × 2 = 6 multiplies the base by the exponent instead of raising it, and gives x = 3.5. The right side is always the base TO the power: 3² = 9.
A negative number on the right: log₂ x = −3
x = 2−3
rewrite as a power
x = 18
positive, so allowed
A negative LOG is fine. Only the argument x has to be positive, and 18 is.
A difference of logs: log₂ (x + 6) − log₂ x = 2
log₂ (x + 6x) = 2
minus means divide
x + 6x = 4
2² = 4
x + 6 = 4x, so x = 2
multiply by x
log₂ 8 − log₂ 2 = 3 − 1 = 2 ✓
check in the original
No solution at all: log(x − 5) = log(2x)
x − 5 = 2x
equal logs, equal arguments
x = −5
the only candidate
log(−10) is undefined
both arguments are negative
no solution
reject it
The algebra always produces a number; only the check can say "no solution".
log₂ x < 4
base 2 is above 1
x < 2⁴ = 16
direction unchanged
and x > 0
from the domain
0 < x < 16
both conditions at once
With a base below 1 the inequality flips, exactly as it did for exponentials. log₍₁ᐟ₂₎ x < 4 gives x > (12)⁴, so x > 116. The domain condition x > 0 is then already covered.
Periodical exams ask you to solve a single log equal to a number, a log equal to a log, or two logs combined into one, and to reject the root that breaks the domain; entrance tests like the version where only one of two roots survives. Combine into one log, rewrite as a power, solve, then test. The 10-second check is to put each root into every argument of the ORIGINAL equation: each must come out positive.
Solve for x
log (x + 3) = 1
No base is written, so it is base 10. Rewriting: x + 3 = 10¹ = 10, so x = 7. Check: log 10 = 1 ✓ A treats the 1 as the answer to x + 3. B subtracts 3 from … nothing that fits. The rule is that the number on the right becomes the EXPONENT on the base.
Hint: "log" with no base written means base 10. Ask what power of 10 the log is telling you about.
