Algebra 3 Quiz: Logarithmic Functions And Graphs

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Logarithmic Functions And GraphsQuestion 1 of 12

The graph of $g$ is the reflection over the line $y=x$ of the graph of $f(x)=3^{x-2}+1$ . Which equation defines $g$ ?

g(x)=\log_3(x+2)-1

g(x)=\log_3(x+1)-2

g(x)=\log_3(x-2)+1

g(x)=\log_3(x-1)+2

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Algebra 3 Quiz

Algebra 3 Quiz: Logarithmic Functions And Graphs

Practice Logarithmic Functions And Graphs in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Logarithmic Functions And Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The graph of $g$ is the reflection over the line $y=x$ of the graph of $f(x)=3^{x-2}+1$ . Which equation defines $g$ ?

$g(x)=\log_3(x+2)-1$
$g(x)=\log_3(x+1)-2$
$g(x)=\log_3(x-2)+1$
$g(x)=\log_3(x-1)+2$ (correct answer)

Explanation: Whenever you see a reflection over the line

y=x

, you are finding the inverse function. That means you swap

x

and

y

and then solve for the new

y

. Start with

y=3^{x-2}+1

, swap to get

x=3^{y-2}+1

. Subtract 1 from both sides:

x-1=3^{y-2}

. Now take the base-3 logarithm of each side:

\log_3(x-1)=y-2

. Finally add 2:

y=\log_3(x-1)+2

. So the correct equation is

g(x)=\log_3(x-1)+2

. Now look at the other choices. The one with

\log_3(x+2)-1

has the numbers swapped and the signs flipped — it looks like someone mistakenly added 2 inside and subtracted 1 outside, which reverses the original shifts incorrectly. The choice

\log_3(x+1)-2

adds 1 inside and subtracts 2 outside — again the constants are in the wrong places, confusing the horizontal shift (from the

-2

in the exponent) with the vertical shift (the

+1

at the end). The choice

\log_3(x-2)+1

keeps the original shifts intact but does not invert them; it treats the function as its own inverse, which is not true for exponential and logarithmic pairs. For a quick check, remember that the inverse of

a^{x-h}+k

\log_a(x-k)+h

. The horizontal shift

h

becomes the vertical shift, and the vertical shift

k

becomes the horizontal shift inside the log. On the exam, always swap and solve — don't just guess based on the original shifts.

Question 2

Let $f(x)=\log_2 x$ and $g(x)=\log_4 x$ . Which statement about their graphs is true?

For $x>1$ , $g(x)<f(x)$ ; for $0<x<1$ , $g(x)<f(x)$ .
For $x>1$ , $g(x)<f(x)$ ; for $0<x<1$ , $g(x)>f(x)$ . (correct answer)
The graph of $g$ is the graph of $f$ shifted $2$ units down.
The graphs are reflections of each other across the $x$ -axis.

Explanation: Whenever you see logarithms with different bases, rewrite them in a common base. Here, use the change-of-base formula:

\log_4 x=\frac{\log_2 x}{\log_2 4}=\frac12\log_2 x.

g

is actually a vertical compression of

f

by a factor of

1/2

, not a shift or reflection. . Now compare the two functions. If

x>1

, then

\log_2 x>0

, so half of a positive number is smaller:

g(x)<f(x)

. But if

0<x<1

, then

\log_2 x<0

, and half of a negative number is closer to zero, so it is greater:

g(x)>f(x)

. For example,

\log_4(1/2)=-1/2

, while

\log_2(1/2)=-1

. Therefore the correct statement is: for

x>1

g(x)<f(x)

; for

0<x<1

g(x)>f(x)

. The first statement claims

g(x)<f(x)

in both intervals. It misses that when both values are negative, "less" means more negative, and g is less negative, so actually greater. The shifted-down claim would require

g(x)=f(x)-2

, but

\log_2 x-2=\log_2(x/4)

, not

\frac12\log_2 x

. The reflection claim would require

g(x)=-\log_2 x

, but the factor here is positive

1/2

, not negative

1

. Study tip: when bases are powers of each other, convert to the same base first. Then test one value in each interval, like

x=2

and

x=1/2

, to see the actual ordering. That quick check avoids both the sign trap and the shift/reflection traps.

Question 3

Let $g$ be the inverse of $f(x)=5^{x-1}$ . What is $g(25)$ ?

$26$
$2$
$3$ (correct answer)
$5$

Explanation: When you see a question about an inverse function, remember that

g(y)

asks: "What input

x

f

produces the output

y

?" So

g(25)

means solving

f(x)=25

. Here,

f(x)=5^{x-1}

, so set

5^{x-1}=25

. Since

25=5^2

, the exponents must match:

x-1=2

, giving

x=3

. Thus

g(25)=3

. Now, why the other choices are traps: choosing

26

comes from mistakenly thinking

g(25)=f(25)+1

, or swapping input/output without solving. Choosing

2

happens if you solve

5^x=25

but forget the

-1

shift in the exponent—that gives

x=2

, but plugging back yields

f(2)=5

, not

25

. Choosing

5

might arise from thinking the base

5

is the answer, or confusing

g(25)

with evaluating

f

25

, which is enormous, not

5

. Your go-to strategy: to find

g(a)

, always write

f(x)=a

and solve for

x

. Watch for transformations like the

-1

inside the exponent—they shift your answer. Practice by rewriting powers as the same base, then equate exponents. This turns inverse problems into simple algebra.

Question 4

Let $f(x)=\log_4(6-2x)+5$ . Which statement about the graph of $f$ is correct?

Domain $(-\infty,3)$ , vertical asymptote $x=3$ , range all real numbers. (correct answer)
Domain $(-\infty,6)$ , vertical asymptote $x=6$ , range all real numbers.
Domain $(3,\infty)$ , vertical asymptote $x=3$ , range $y>5$ .
Domain $(-\infty,3)$ , vertical asymptote $x=3$ , range $y>5$ .

Explanation: When you see a logarithmic function like

f(x)=\log_4(6-2x)+5

, think about the two conditions that define its graph: the argument of the log must be positive, and the log itself can output any real number. The domain comes from solving

6-2x>0

, which gives

-2x>-6

, so

x<3

. Thus the domain is

(-\infty,3)

, and the vertical asymptote occurs where the argument equals zero:

6-2x=0

, so

x=3

. The graph approaches this asymptote but never touches it. As for range, a logarithm alone has range all real numbers, and adding

5

only shifts the graph vertically, so the range remains all real numbers. That confirms the correct statement: domain

(-\infty,3)

, vertical asymptote

x=3

, range all real numbers. Now look at the traps. The choice with domain

(-\infty,6)

and asymptote

x=6

comes from forgetting the coefficient

-2

and solving

6-x>0

incorrectly; the asymptote is tied to where the inside becomes zero, which is

x=3

, not

6

. The choice with domain

(3,\infty)

reverses the inequality sign, a common error when dividing by a negative number — remember

x<3

, not

x>3

. Finally, the choice with domain

(-\infty,3)

and asymptote

x=3

but range

y>5

confuses a logarithmic function with an exponential or square-root function; logs are not bounded below, so the range is not restricted to values above the vertical shift. Your takeaway: for any logarithmic function, set the argument greater than zero for the domain, set it equal to zero for the vertical asymptote, and remember the range is always all real numbers. Watch the sign when solving inequalities involving a negative coefficient.

Question 5

Let $f(x)=-2\log_3(x+4)$ . Which statement about the graph of $f$ is true?

The $x$ -intercept is $-1$ , and $f(x)\to -\infty$ as $x\to -4^+$ .
The $x$ -intercept is $-3$ , and $f(x)\to -\infty$ as $x\to -4^+$ .
The $x$ -intercept is $-1$ , and $f(x)\to +\infty$ as $x\to -4^+$ .
The $x$ -intercept is $-3$ , and $f(x)\to +\infty$ as $x\to -4^+$ . (correct answer)

Explanation: When you see a transformed logarithmic function like this, start with two anchor points: the vertical asymptote and the

x

-intercept. For

f(x)=-2\log_3(x+4)

, the domain is

x>-4

, so the vertical asymptote is

x=-4

. To find the

x

-intercept, set

f(x)=0

-2\log_3(x+4)=0

, so

\log_3(x+4)=0

, meaning

x+4=1

, hence

x=-3

. That eliminates any statement claiming the

x

-intercept is

-1

. Now examine the behavior near the asymptote. As

x\to -4^+

x+4\to 0^+

, so

\log_3(x+4)\to -\infty

. Multiplying by

-2

flips the sign, giving

f(x)\to +\infty

. Therefore the true statement is the one with

x

-intercept

-3

and

f(x)\to +\infty

x\to -4^+

. Why the others fail: the choices with

x

-intercept

-1

confuse the zero of the logarithm with

x+4=3

, but

\log_3(3)=1

, not

0

. The choice with

x

-intercept

-3

but

f(x)\to -\infty

correctly finds the intercept but forgets that the leading coefficient

-2

reverses the direction of the unbounded behavior. Study tip: always separate intercept-finding from end-behavior analysis. For logarithms, set the inside equal to

1

for the

x

-intercept, and use the sign of the leading coefficient to decide whether the graph rises or falls into the vertical asymptote.

Question 6

The graph of $f(x)=b^x$ , where $b>1$ , has the $x$ -axis as a horizontal asymptote. Which statement best explains why the graph of $f^{-1}(x)=\log_b x$ has a vertical asymptote at $x=0$ ?

The value $\log_b(0)$ is undefined, so the graph cannot touch or cross the $y$ -axis; it must therefore have a vertical asymptote there.
Reflecting $f$ across $y=x$ sends $y=0$ to $x=0$ ; the range $(0,\infty)$ of $f$ is the domain of $f^{-1}$ . (correct answer)
Since the range of $f^{-1}$ is all real numbers, the graph cannot level off; its only possible asymptote is the vertical line $x=0$ .
Because $b^0=1$ ,the inverse has $x$ -intercept $1$ ;the gap between that intercept and the $y$ -axis forces a vertical asymptote at $x=0$ .

Explanation: When you see a question about inverse functions and asymptotes, remember the central idea: reflecting a graph across the line

y=x

swaps the roles of

x

and

y

. That reflection also swaps domain and range, and it turns horizontal lines into vertical lines. For

f(x)=b^x

b>1

, the domain is all real numbers, but the range is

(0,\infty)

. The graph approaches the

x

-axis,

y=0

, as

x\to -\infty

, so

y=0

is a horizontal asymptote. Its inverse,

f^{-1}(x)=\log_b x

, must have domain

(0,\infty)

, exactly the range of

f

. Reflecting across

y=x

sends the horizontal line

y=0

to the vertical line

x=0

, which is why the logarithmic graph has a vertical asymptote at

x=0

. That is the reasoning behind the correct statement: reflecting

f

sends

y=0

x=0

, and the range of

f

becomes the domain of

f^{-1}

. Now look at the distractors. Saying

\log_b(0)

is undefined and therefore there must be a vertical asymptote confuses "undefined at a point" with "unbounded behavior near that point"; a function can be undefined at 0 without having an asymptote. Saying the range of

f^{-1}

is all real numbers explains why there is no horizontal asymptote, but it does not force a vertical asymptote at 0. And because

b^0=1

, the

x

-intercept is 1, but a gap between that intercept and the

y

-axis does not create an asymptote; many graphs have an

x

-intercept at 1 with no vertical asymptote. Study tip: for inverse functions, always translate asymptotes across

y=x

: a horizontal asymptote of

f

becomes a vertical asymptote of

f^{-1}

Question 7

Let $f(x)=\log_3(x-6)$ . Which statement about the graph of $f^{-1}$ is correct?

Its domain is $(6,\infty)$ , its range is all real numbers, and it has vertical asymptote $x=6$ .
Its domain is all real numbers, its range is $(6,\infty)$ , and it has horizontal asymptote $y=6$ . (correct answer)
Its domain is all real numbers, its range is $(0,\infty)$ , and it has horizontal asymptote $y=0$ .
Its domain is all real numbers, its range is $(6,\infty)$ , and it has vertical asymptote $x=6$ .

Explanation: When you see a question about the graph of an inverse function, remember that the inverse swaps the domain and range of the original function, and its graph is a reflection across the line

y=x

. For

f(x)=\log_3(x-6)

, the logarithm is defined only when

x-6>0

, so the domain of

f

(6,\infty)

, and its range is all real numbers. Therefore, the inverse

f^{-1}

has domain all real numbers and range

(6,\infty)

. Also, because

f

has a vertical asymptote at

x=6

, its reflection has a horizontal asymptote at

y=6

. That matches the statement giving domain all real numbers, range

(6,\infty)

, and horizontal asymptote

y=6

. The statement claiming domain

(6,\infty)

, range all real numbers, and vertical asymptote

x=6

actually describes the original function

f

, not its inverse. The statement with range

(0,\infty)

and horizontal asymptote

y=0

confuses the inverse with a basic exponential graph like

y=3^x

, but the correct inverse here is

y=3^x+6

, whose range starts at 6. The statement with domain all real numbers, range

(6,\infty)

, but vertical asymptote

x=6

gets the domain and range right but misidentifies the asymptote: an inverse of a vertical asymptote is a horizontal asymptote, not another vertical one. A useful habit: whenever you find an inverse, immediately swap the domain and range, and swap "vertical" and "horizontal" in any asymptote description.

Question 8

The graph of $f(x)=\log_2 x$ is reflected across the $x$ -axis and then shifted 3 units to the right. Which function has the resulting graph?

$h(x)=-\log_2(x+3)$
$h(x)=\log_2(3-x)$
$h(x)=-\log_2(x-3)$ (correct answer)
$h(x)=-\log_2 x+3$

Explanation: When you see transformations of a function like

f(x)=\log_2 x

, always track two things: the order of operations and where each change acts. Reflections and shifts outside the function (like multiplying by –1 or adding a constant) affect the output; changes inside the argument (like replacing

x

with

x-3

) affect the input. Here, "reflected across the x-axis" means you multiply the entire output by –1, giving

-\log_2 x

. Then "shifted 3 units to the right" means every input is replaced by

x-3

, so the result is

-\log_2(x-3)

. The choice

-\log_2(x+3)

would be a reflection across the x-axis but a shift left 3 units, not right. The choice

\log_2(3-x)

is a reflection across the y-axis (because the negative sign is inside the argument) combined with a right shift, not an x-axis reflection. The choice

-\log_2 x + 3

reflects across the x-axis but then shifts up 3 units, not right. Only

-\log_2(x-3)

applies the two transformations in the correct order and direction. A useful memory aid: for horizontal shifts, the sign inside the argument is opposite to the direction—

x-3

moves right,

x+3

moves left. And a negative sign outside flips vertically; a negative sign inside flips horizontally. On the exam, write out each transformation step by step, and always check whether the shift is inside or outside the function.

Question 9

The point $(5,1)$ lies on the graph of $f(x)=\log_2(x+3)-2$ . If $g$ is the inverse of $f$ , which point must lie on the graph of $g$ ?

$(1,5)$ (correct answer)
$(5,1)$
$(1,-5)$
$(5,-1)$

Explanation: Whenever you see a question about an inverse function, think about the defining relationship: if

(a,b)

lies on

f

, then

(b,a)

must lie on

f^{-1}

. An inverse swaps the input and output, which geometrically reflects the graph across the line

y=x

. Since

(5,1)

lies on

f

, the inverse

g

must contain the swapped point

(1,5)

. That is exactly the point that must lie on the graph of

g

. The choice

(5,1)

is just the original point on

f

, not on

g

— it would mean the function and its inverse share the same coordinate pair, which isn't implied here. The choice

(1,-5)

might come from incorrectly subtracting the

2

from the

y

-value or misapplying the logarithm's vertical shift; no swapping of coordinates is involved. The choice

(5,-1)

also swaps incorrectly, perhaps by negating the

y

-coordinate or confusing the

-2

outside the logarithm with the output value. None of those reflect the actual inverse rule. A quick strategy for inverse questions: whenever you are given a point on a function and asked for a point on its inverse, simply switch the coordinates. No calculation with the logarithm formula is needed. Watch out for traps that keep the same point or alter the sign — the inverse only swaps the ordered pair, nothing else.

Question 10

Let $f(x)=\log_{1/3}(x-2)$ . Which statement best describes the graph of $f$ ?

It has horizontal asymptote $y=2$ , and $f(x)\to -\infty$ as $x\to 2^+$ .
It has vertical asymptote $x=2$ , and $f(x)\to +\infty$ as $x\to\infty$ .
It has vertical asymptote $x=2$ , and $f(x)\to -\infty$ as $x\to\infty$ . (correct answer)
It has vertical asymptote $x=-2$ , and $f(x)\to +\infty$ as $x\to\infty$ .

Explanation: Whenever you see a logarithmic function, first find the domain by setting the inside greater than

0

. That single step gives you the vertical asymptote. Here

f(x)=\log_{1/3}(x-2)

, so the inside is

x-2

, meaning

x>2

and the asymptote is

x=2

, not

-2

. Since the base

1/3

is between

0

and

1

, this log function is decreasing: as each input grows, the output becomes more negative. So as

x\to\infty

, we have

x-2\to\infty

, and

f(x)\to -\infty

. The statement that combines vertical asymptote

x=2

with

f(x)\to -\infty

x\to\infty

is correct. Now look at the distractors. The statement saying there is a horizontal asymptote

y=2

is a red herring: logarithmic functions do not level off to a horizontal asymptote; they diverge as the input grows. That same choice also claims

f(x)\to -\infty

x\to 2^+

, but near a vertical asymptote with base

<1

, the values actually rush upward to

+\infty

. The statement with vertical asymptote

x=2

and

f(x)\to +\infty

x\to\infty

correctly identifies the asymptote but reverses the end behavior, since the decreasing log base sends large inputs down to

-\infty

. Finally, the statement with vertical asymptote

x=-2

is wrong about the location

x=2

and repeats the wrong end behavior. Study tip: For any log, set the argument equal to

0

to find the vertical asymptote. Then compare the base to

1

: base

>1

means increasing, base

<1

means decreasing. That pattern tells you the behavior at infinity quickly and reliably.

Question 11

Let $b>1$ . Compared with the graph of $f(x)=\log_b x$ , the graph of $g(x)=\log_{1/b} x$ is which of the following?

The reflection of $f$ across the $x$ -axis, with the same domain and $x$ -intercept. (correct answer)
The reflection of $f$ across the $y$ -axis, with domain $(-\infty,0)$ and $x$ -intercept $-1$ .
The reflection of $f$ across the line $y=x$ , so $g$ is the inverse of $f$ .
The same graph as $f$ , because every logarithmic function passes through $(1,0)$ .

Explanation: When you see a logarithm with a fractional base, your first instinct should be to rewrite it using the exponent property of logarithms. Since

1/b = b^{-1}

, we have

\log_{1/b} x = \log_{b^{-1}} x = -\log_b x

. This means

g(x) = -f(x)

. Multiplying a function by

-1

flips it vertically across the

x

-axis. The domain of any logarithmic function

\log_a x

(0, \infty)

, regardless of the base

a

, and the

x

-intercept is always at

x=1

because

\log_a 1 = 0

. Therefore,

g

has the exact same domain and

x

-intercept as

f

, but its

y

-values are negated. Now look at the other choices. The reflection across the

y

-axis would produce

\log_b (-x)

, which has a domain of

(-\infty, 0)

and an

x

-intercept at

-1

; that is a different function, not what we have here. The reflection across the line

y=x

would give the inverse, which is

b^x

, an exponential function, not another logarithm. The choice claiming they are the same graph is tempting because both pass through

(1,0)

, but they are not identical: for

x>1

f(x)>0

while

g(x)<0

(since

b>1

f

is increasing,

g

is decreasing). Study tip: Memorize the identity

\log_{1/b} x = -\log_b x

. Recognizing this instantly saves you time and prevents confusion about domain or intercepts—a reciprocal base only causes a vertical reflection, never a change in domain.

Question 12

Which of the following equations, all with base 3, could define a function whose graph has vertical asymptote $x=2$ , contains $(3,0)$ , and is decreasing on its domain?

$f(x)=-\log_3(2-x)$
$f(x)=\log_3(2-x)$
$f(x)=\log_3(x-2)$
$f(x)=-\log_3(x-2)$ (correct answer)

Explanation: Whenever you see a transformed logarithmic function, ask three things: Where does the argument become

0

? That gives the vertical asymptote. What is the domain? That tells you whether a stated point can exist. And what is the sign of the coefficient? With base

3>1

, a negative coefficient makes the function decreasing. For an asymptote at

x=2

, the argument must become zero at

x=2

, so both forms with

2-x

and

x-2

are candidates. The choices

\log_3(2-x)

and

-\log_3(2-x)

have domain

x<2

, so

x=3

is not in their domain; they cannot contain

(3,0)

. Also,

-\log_3(2-x)

is actually increasing on its domain, so it fails the decreasing condition as well. Now consider

\log_3(x-2)

. It has domain

x>2

, asymptote

x=2

, and at

x=3

gives

\log_3(1)=0

, so it contains

(3,0)

. But its derivative is positive, so it increases. The negative version

-\log_3(x-2)

flips it: same asymptote and same point because

-\log_3(1)=0

, but its derivative is negative, so it decreases. Thus it matches all conditions. Study tip: for

y=\pm\log_b(x-h)

, the asymptote is

x=h

, and with

b>1

, decreasing requires the negative sign. Always test the given point by setting the argument to

1