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Exponential Functions and Graphs: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Three powers

Evaluate $16^0+16^{-1}+16^{1/2}$ .
1. Hint 1
  Zero, negative, and fractional exponents each have a meaning fixed by the exponent laws.
2. Hint 2
  The three terms become $1$ , the reciprocal of $16$ , and its positive square root.
Answer

$\frac{81}{16}$ .

Full solution

The base is positive, so each rule applies.
$\begin{aligned}&16^0+16^{-1}+16^{1/2}\\&=1+\frac1{16}+4\end{aligned}$
Adding with denominator $16$ gives $\frac{81}{16}$ .

Answer

$\frac{81}{16}$ .

Key idea

Exponent rules determine zero, negative, and fractional powers of a positive base.
Problem 1: Mark complete
Problem 2 A fractional input

For $f(x)=27^{2x}$ , evaluate $f(-\frac16)$ .
1. Hint 1
  Substitute the input into the whole exponent.
2. Hint 2
  A negative one-third power is the reciprocal of a cube root.
Answer

$\frac13$ .

Full solution

The exponent is $2(-\frac16)=-\frac13$ .
$f(-\tfrac16)=27^{-1/3}$
The positive cube root of $27$ is $3$ , so the reciprocal is $\frac13$ .
Its cube is $\frac1{27}$ , as required.

Answer

$\frac13$ .

Key idea

A coefficient in an exponent acts on the input before the power is evaluated.
Problem 2: Mark complete
Problem 3 A combined output

Solve $4^{x+1}+4^x=80$ for real $x$ .
1. Hint 1
  The two terms contain the same power of $4$ .
2. Hint 2
  Rewrite $4^{x+1}$ as $4\cdot4^x$ and collect.
Answer

$x=2$ .

Full solution

Collect the common exponential factor.
$5\cdot4^x=80$
Divide by $5$ , giving $4^x=16$ .
Since $16=4^2$ and base $4$ is one-to-one, $x=2$ .

Check: the original left side is $64+16$ , which is $80$ .

Answer

$x=2$ .

Key idea

Like exponential terms can be collected before applying the one-to-one property.
Problem 3: Mark complete
Problem 4 An incomplete record

A record has inputs $-1,0,1,2$ and outputs $12,6,m,\frac32$ , in that order. Find $m$ and a rule $f(x)=ab^x$ that fits every entry. Show that the neighboring output ratios agree.
1. Hint 1
  The inputs are equally spaced, so an exponential requires equal ratios.
2. Hint 2
  The ratio from the first known output to the next fixes the missing output.
Answer

$m=3$ ; $f(x)=6(\frac12)^x$ ; each ratio is $\frac12$ .

Full solution

The known first ratio is
$6/12=\frac12.$
Thus
$m=6\cdot\frac12$
so $m=3$ .
The remaining ratios are $3/6=\frac12$ and
$(\frac32)/3=\frac12.$

The output at zero is $6$ , so $a=6$ , and the unit-step ratio gives $b=\frac12$ .
The rule gives $12,6,3,\frac32$ at the four inputs.

Answer

$m=3$ ; $f(x)=6(\frac12)^x$ ; each ratio is $\frac12$ .

Key idea

In an exponential table with equally spaced inputs, a missing interior output must keep the common ratio at both of its neighboring steps.
Problem 4: Mark complete
Problem 5 Two curves to compare

On the blank axes in the figure, sketch $p(x)=(\frac25)^x$ and $g(x)=3-(\frac25)^{x-1}$ . Label each curve and its horizontal asymptote. State the domain, range, and direction of each.

Blank axes for your sketch.

Text description of this figure

An empty grid for sketching: the x-axis runs from -3 to 4 and the y-axis from -2 to 5, on equal unit scales, with a grid line and a label at every integer. No curve, asymptote or point is drawn.
1. Hint 1
  Use easy integer inputs to place points before drawing each curve.
2. Hint 2
  Track the horizontal shift, the reflection, and the vertical shift in $g$ .
Answer

$p$ : domain $\mathbb R$ , range $y>0$ , decreasing, asymptote $y=0$ . $g$ : domain $\mathbb R$ , range $y<3$ , increasing, asymptote $y=3$ . The sketches are in the full solution.

Full solution

For $p$ , the points $(-1,\frac52)$ , $(0,1)$ , and $(1,\frac25)$ place a decreasing curve above $y=0$ .
The exponential is defined for every real input, and its outputs are all positive.

For $g$ , the term $(\frac25)^{x-1}$ is positive and decreasing, so subtracting it from $3$ gives an increasing function whose outputs stay below $3$ .
It passes through $(0,\frac12)$ , $(1,2)$ , and $(2,\frac{13}5)$ , with domain all reals, range $y<3$ , and asymptote $y=3$ , approached from below as $x$ grows.

The sketches of p and g.

Answer

$p$ : domain $\mathbb R$ , range $y>0$ , decreasing, asymptote $y=0$ . $g$ : domain $\mathbb R$ , range $y<3$ , increasing, asymptote $y=3$ . The sketches are in the full solution.

Key idea

A reflection across the x-axis turns a decreasing exponential into an increasing one, and a vertical shift moves its asymptote by the same amount.
Problem 5: Mark complete
Problem 6 A smaller input step

A function $f(x)=ab^x$ has $f(0)=10$ . Every increase of $\frac12$ in the input multiplies its output by $3$ . Find $a$ , $b$ , and $f(-1)$ .
1. Hint 1
  Two half-steps make one unit step.
2. Hint 2
  To move backward a unit, undo the factor for a whole unit.
Answer

$a=10$ , $b=9$ , $f(-1)=\frac{10}{9}$ .

Full solution

The zero input gives $a=10$ .
Two multiplications by $3$ give the unit-step multiplier.
$b=3^2$
Thus $b=9$ and
$f(x)=10\cdot9^x.$

The output one unit before zero is $10/9$ .
Advancing two half-steps multiplies this by $3$ twice, returning $10$ .

Answer

$a=10$ , $b=9$ , $f(-1)=\frac{10}{9}$ .

Key idea

A whole-step multiplier is built by multiplying the factors of its equal smaller steps.
Problem 6: Mark complete
Problem 7 A shifted record

A function has the form $g(x)=a2^x+k$ . Its outputs at $x=0$ and $x=1$ are $7$ and $11$ . Find its rule, horizontal asymptote, and output at $x=-1$ .
1. Hint 1
  Both recorded outputs contain the same vertical shift.
2. Hint 2
  Write $a+k=7$ and $2a+k=11$ , then subtract to eliminate $k$ .
Answer

$g(x)=4\cdot2^x+3$ ; asymptote $y=3$ ; $g(-1)=5$ .

Full solution

The data gives $a+k=7$ and $2a+k=11$ .
Subtracting gives $a=4$ , and then $k=3$ .
$g(x)=4\cdot2^x+3$

The exponential term approaches zero to the left, giving asymptote $y=3$ .
Also $g(-1)=4(\frac12)+3$ , which is $5$ .
The rule returns both stated outputs.

Answer

$g(x)=4\cdot2^x+3$ ; asymptote $y=3$ ; $g(-1)=5$ .

Key idea

An additive shift can be separated from an exponential coefficient by comparing two outputs.
Problem 7: Mark complete
Problem 8 Choosing a parameter

A designer proposes $f(x)=(c-2)^x$ as a nonconstant real exponential defined for every real $x$ . Which real values of $c$ are allowed? Explain why the excluded boundary and constant cases fail.
1. Hint 1
  Check the base, which is the entire expression $c-2$ .
2. Hint 2
  Test the base at a fractional input and at a negative input, and consider what base 1 does.
Answer

$c>2$ and $c\ne3$ .

Full solution

The base condition $c-2>0$ gives $c>2$ .
The exclusion of base one gives $c\ne3$ .
$c-2\ne1$

If $c<2$ , the input $x=\frac12$ asks for $(c-2)^{1/2}$ , the square root of a negative number, which is not real.
If $c=2$ , the input $-1$ demands division by zero.
If $c=3$ , every output is $1$ , so the function is constant.

Answer

$c>2$ and $c\ne3$ .

Key idea

Restrictions on an exponential base apply to the whole expression used as that base.
Problem 8: Mark complete
Problem 9 A proposed identity

Let $f(x)=11\cdot19^x$ . Is $f(s+t)=f(s)f(t)$ true for all real $s,t$ ? If it fails, replace the right side with a constant multiple of $f(s)f(t)$ that makes it true.
1. Hint 1
  The exponential factors obey the exponent law, but the leading coefficients also multiply.
2. Hint 2
  Compare the coefficient of $19^{s+t}$ on the two sides.
Answer

No; $f(s+t)=\frac1{11}f(s)f(t)$ .

Full solution

Multiplying the two outputs gives
$f(s)f(t)=121\cdot19^{s+t}$
whereas
$f(s+t)=11\cdot19^{s+t}.$
Therefore multiplying the product by $\frac1{11}$ repairs the identity.

For example, at $s=t=0$ the proposed equality would say $11=121$ , so it fails.
The corrected right side is $121/11$ , which is $11$ .

Answer

No; $f(s+t)=\frac1{11}f(s)f(t)$ .

Key idea

For $f(x)=ab^x$ with $a\ne0$ and $b>0$ , the product law becomes $f(s+t)=\frac{f(s)f(t)}a$ .
Problem 9: Mark complete
Problem 10 One base, three behaviors

For a real constant $c$ , let $H_c(x)=(c-1)\cdot13^x+c$ . A student claims every such graph increases because $13>1$ . Classify the direction and range of $H_c$ for every $c$ , and decide whether the claim is correct.
1. Hint 1
  The sign of the coefficient of $13^x$ decides the direction.
2. Hint 2
  Consider separately a positive, a negative, and a zero coefficient.
Answer

The claim is false. For $c>1$ : increasing, range $y>c$ . For $c<1$ : decreasing, range $y<c$ . For $c=1$ : constant, range $\{1\}$ .

Full solution

The term $13^x$ is positive and increasing, and takes every positive value.
Multiplying it by $c-1$ keeps it increasing when $c-1>0$ , reverses it when $c-1<0$ , and makes it zero when $c=1$ .

If $c>1$ , then $(c-1)13^x$ takes every positive value, so $H_c$ increases with range $y>c$ .
If $c<1$ , it takes every negative value, so $H_c$ decreases with range $y<c$ .
If $c=1$ , then $H_1(x)=1$ for every $x$ , a constant function with range $\{1\}$ .

So the base alone does not decide the direction: the sign of the coefficient does, and the claim fails for every $c<1$ and at $c=1$ .

Answer

The claim is false. For $c>1$ : increasing, range $y>c$ . For $c<1$ : decreasing, range $y<c$ . For $c=1$ : constant, range $\{1\}$ .

Key idea

For $ab^x+k$ with $b>1$ , the graph increases when $a>0$ , decreases when $a<0$ , and is constant when $a=0$ .
Problem 10: Mark complete

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