The log graph is the exponential seen in a mirror.
The exponential's (0, 1) becomes the logarithm's (1, 0). Its horizontal asymptote becomes a vertical one. Every feature turns on its side.
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- Graphing Logarithmic Functions
Graphing Logarithmic Functions
Reflect the exponential in y = x and everything swaps: the point, the asymptote, the domain.
Find the vertical asymptote
y = log (x − 2)
A logarithm is undefined when its argument reaches 0, so the wall sits where x − 2 = 0, which is x = 2. The domain is x > 2. A is the asymptote before the shift. D flips the sign — inside the bracket, "− 2" moves the graph two to the RIGHT, not left. C is horizontal, and log graphs have no horizontal asymptote at all.
Domain positive, range everything.
y = log₂ x
domain x > 0
no logs of 0 or below
range
all real numbers
outputs go as low as you like
x-intercept
(1, 0)
and no y-intercept
log₂(0.001) ≈ −10 and log₂(0.000001) ≈ −20. The curve falls forever as x approaches 0 from the right, which is what the vertical asymptote describes.
The domain is wherever the argument is positive.
y = log(x − 2)
argument is x − 2
x − 2 > 0
the only condition
x > 2
asymptote at x = 2
Finding the domain and finding the asymptote are the same piece of work. Solve the argument > 0 and both answers fall out.
Inside the bracket moves sideways; outside moves up.
y = log(x − 2)
right 2
asymptote x = 2
y = log(x + 5)
left 5
asymptote x = −5
y = log x + 3
up 3
asymptote still x = 0
A shift inside goes the opposite way to the sign
"− 2" moves the graph two to the right. The reason is plain if you ask which x makes the bracket behave as x = 0 did before: x = 2. Reading the minus as "left" is the single most common error on this topic.
When x is subtracted inside the log.
The rule does not change — the argument must be positive — but solving the inequality now needs care, because x carries a minus sign.
Find the domain, asymptote and x-intercept of y = log(7 − x).
7 − x > 0
the argument must be positive
7 > x, so x < 7
add x to both sides
asymptote x = 7
where the argument is 0
7 − x = 1, so x = 6: (6, 0)
log is 0 when the argument is 1
Check x = 0, inside the domain: log 7 is defined ✓. The graph lives to the LEFT of its wall and falls toward it.
Dividing by −1 flips the sign
Solving 7 − x > 0 as −x > −7 and then writing x > −7, or x > 7, keeps the arrow pointing the wrong way. Dividing by −1 turns > into <. Moving x to the right side avoids the trap entirely.
Variations: a multiplier inside, a shift outside, a base below 1.
A multiplier inside: y = log₃ (3x)
3x = 1, so x = 13
x-intercept (13, 0)
asymptote x = 0
3x = 0 still at 0
log₃ (3x) = 1 + log₃ x
the product law
A multiplier inside acts like a shift UP by its log: this is y = log₃ x lifted by 1.
A shift outside: y = log₂ x − 3
log₂ x − 3 = 0
x-intercept: set y = 0
log₂ x = 3, so x = 2³ = 8
(8, 0)
asymptote still x = 0
an outside shift leaves the wall alone
An outside shift moves the x-intercept away from (1, 0) but not the asymptote.
A base below 1: y = log₍₁ᐟ₂₎ x
(1, 0)
log of 1 is still 0
(2, −1) and (12, 1)
(12)−1 = 2 and (12)¹ = 12
the curve falls left to right
a decreasing log
Same domain x > 0 and same wall x = 0, but the graph is the rising one turned upside down.
The asymptote and one point are enough.
asymptote: x = 3
where x − 3 = 0
x-intercept: x − 3 = 1
log is 0 when the argument is 1
(4, 0)
plot it
(5, 1) since log₂ 2 = 1
one more point fixes the shape
On the exam
Periodical exams ask for the domain, the vertical asymptote, the x-intercept or the matching exponential graph; entrance tests add a shift, or put the x after a minus sign, as in log(4 − x). Set the argument greater than 0 for the domain and asymptote, and equal to 1 for the x-intercept. For a 10-second check, pick one x inside your domain and confirm the argument comes out positive.
Now you try
Practice problem
Find the domain
y = log (x − 5)
The argument must be strictly positive: x − 5 > 0, so x > 5. B includes 5 itself, where the argument is 0 and the log is undefined — a log graph never touches its asymptote. C has the inequality backwards.
Hint: set the bracket greater than zero, not greater than or equal. Zero is exactly the value a logarithm cannot take.