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Exponential Functions: 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 The input instruction

A rule adds $2$ to its real input $x$ , raises the fixed number $3$ to that sum, and halves the value obtained. Write the rule in the form $f(x)=a\cdot b^x$ .
1. Hint 1
  The entered quantity becomes an exponent rather than a base.
2. Hint 2
  Split the power at the addition in its exponent, then apply the halving to the constant factor that appears.
Answer

$f(x)=\tfrac92\cdot3^x$ , or $4.5\cdot3^x$ .

Full solution

The instructions give
$f(x)=\tfrac12\cdot3^{x+2}$
The exponent rule separates the added part:
$3^{x+2}=9\cdot3^x$

Halving that constant gives
$f(x)=\tfrac92\cdot3^x$
so $a=\tfrac92$ and $b=3$ .
At $x=0$ the instructions give $\tfrac12\cdot3^2=\tfrac92$ , which matches.

Answer

$f(x)=\tfrac92\cdot3^x$ , or $4.5\cdot3^x$ .

Key idea

An addition inside an exponent becomes a constant multiplier, which then combines with whatever multiplier sits outside the power.
Problem 1: Mark complete
Problem 2 The base setting

A device uses $f(x)=(k-2)^x$ as a genuine exponential function, defined for every real $x$ and not a constant. Find all allowed real settings $k$ .
1. Hint 1
  The proposed base must meet both restrictions for a real exponential function.
2. Hint 2
  An exponential base must be positive and must not equal $1$ ; apply each restriction to the expression $k-2$ .
Answer

$k>2$ , with $k\ne3$ .

Full solution

Positivity of the base requires
$k-2>0$
or $k>2$ .
The base is not allowed to equal $1$ , so
$k-2\ne1$
excludes $k=3$ .

Every remaining setting supplies a positive base different from $1$ , so the function is defined at all real inputs and is not constant.

Answer

$k>2$ , with $k\ne3$ .

Key idea

A parameter used inside an exponential base must satisfy the base restrictions after its expression is evaluated.
Problem 2: Mark complete
Problem 3 The two stored values

For $f(x)=b^x$ , where $b>0$ and $b\ne1$ , two real inputs satisfy $f(u)=7$ and $f(v)=4$ . Find $f(u-v)$ .
1. Hint 1
  Subtracting inputs corresponds to dividing powers of the same base.
2. Hint 2
  The denominator is positive, so the quotient $b^u/b^v$ is defined.
Answer

$\frac74$ .

Full solution

Since $b^v=4\ne0$ , the quotient rule gives
$b^{u-v}=\frac{b^u}{b^v}$

Substituting the stored outputs gives
$f(u-v)=\frac74$
Multiplying this by $f(v)=4$ recovers $f(u)=7$ , which checks the result.

Answer

$\frac74$ .

Key idea

For a base power, an input difference is represented by a quotient of positive outputs.
Problem 3: Mark complete
Problem 4 The two marked points

The graph shows $f(x)=b^x$ for an allowed base, with two points marked. Read $b$ from the graph. Then say whether $f(-3)$ is greater or less than $1$ , and justify that comparison.

The graph of $f(x)=b^x$ , with two points marked.

Text description of this figure

A coordinate grid with equal unit spacing on both axes. The horizontal axis runs from negative 2 to 2 and the vertical axis from 0 to 5, with gridlines, ticks and labels at every whole number. A smooth curve rises from left to right: it enters at the left just above the horizontal axis, climbs gently at first and then steeply, and leaves through the top of the frame between the inputs 1 and 2. An arrowhead at each end shows that the curve continues. Two points on the curve are marked with filled dots and no printed coordinates: one where the curve crosses the vertical axis, 1 unit above the origin, and one above the input 1, 3 units above the horizontal axis.
1. Hint 1
  Each marked point pairs an input with the power of $b$ that it produces.
2. Hint 2
  The height at input $1$ is $b$ itself, and a negative input turns that power into a reciprocal.
Answer

$b=3$ ; $f(-3)=\tfrac1{27}$ , which is less than $1$ .

Full solution

The marked point above input $1$ has height $3$ , and $f(1)=b$ , so
$b=3$

The asked input lies outside the drawn window, so use the rule rather than the picture.
A negative exponent gives a reciprocal:
$f(-3)=\frac1{3^3}$
So $f(-3)=\tfrac1{27}$ , which is less than $1$ .

The graph agrees with this.
The curve rises from left to right through the other marked point $(0,1)$ , so every input to the left of $0$ produces an output below $1$ .

Answer

$b=3$ ; $f(-3)=\tfrac1{27}$ , which is less than $1$ .

Key idea

A growth exponential passes through $(0,1)$ and stays below $1$ at every negative input, where the power becomes a reciprocal.
Problem 4: Mark complete
Problem 5 The revised time unit

A model is $N(t)=18\cdot3^t$ , where $t$ is time in hours. Another display uses $s$ , the number of half-hours since the same start. Write the same model as $N=a\cdot c^s$ , and state the exact reading when $s=3$ .
1. Hint 1
  Changing the time unit changes how many input steps fit in an hour.
2. Hint 2
  One hour contains two half-hours, so substitute $t=s/2$ in the exponent.
Answer

$N=18(\sqrt3)^s$ ; at $s=3$ , $N=54\sqrt3$ , about $93.5$ .

Full solution

The time inputs satisfy $t=s/2$ .
Substitute this into the original rule:
$N=18\cdot3^{s/2}$

The power rule writes the same function as
$N=18(\sqrt3)^s$
Two steps in $s$ multiply the output by $(\sqrt3)^2=3$ , matching one hour.

At $s=3$ the cube of $\sqrt3$ is $3\sqrt3$ , so
$N=54\sqrt3$
about $93.5$ .
That input is $t=1.5$ hours, and $18\cdot3^{1.5}$ gives the same reading.

Answer

$N=18(\sqrt3)^s$ ; at $s=3$ , $N=54\sqrt3$ , about $93.5$ .

Key idea

When input units change, the exponential base changes so the multipliers over equal physical intervals still agree.
Problem 5: Mark complete
Problem 6 The display offset

A reserve follows $R(t)=ab^t$ , with $a>0$ and $0<b<1$ , where $t$ counts days. A display adds a fixed offset of $6$ units to every true amount. Its readings at days $0$ , $1$ , and $2$ are $70$ , $38$ , and $22$ units. Find the true rule and the display reading at day $3$ .
1. Hint 1
  Remove the fixed offset before looking for the multiplying pattern.
2. Hint 2
  Compare the true amounts at equal day steps, then restore the offset only after predicting a true amount.
Answer

$R(t)=64(\frac12)^t$ ; day $3$ display: $14$ units.

Full solution

Removing the offset gives true amounts $64$ , $32$ , and $16$ units.
Both ratios equal one half, so
$R(t)=64\left(\frac12\right)^t$

The true amount at day $3$ is
$R(3)=8$
units.
Restoring the display offset gives
$8+6=14$
units.

Adding the offset back to $64$ , $32$ , and $16$ reproduces the readings $70$ , $38$ , and $22$ , so the rule fits.

Answer

$R(t)=64(\frac12)^t$ ; day $3$ display: $14$ units.

Key idea

A fixed display offset must be removed before testing an exponential ratio and restored after predicting the underlying amount.
Problem 6: Mark complete
Problem 7 The two counters

For whole-number inputs $n\ge0$ , counter A reports $2\cdot3^n$ and counter B reports $10+4n$ . Find the first input at which A exceeds B. Explain why seeing B larger at the start does not contradict exponential growth eventually outgrowing linear growth.
1. Hint 1
  Compare the two actual outputs at successive whole-number inputs.
2. Hint 2
  One counter multiplies by a fixed factor while the other adds a fixed amount.
Answer

$n=3$ ; A reports $54$ and B reports $22$ .

Full solution

At $n=0$ , A and B give $2$ and $10$ .
At $n=1$ , they give $6$ and $14$ .
At $n=2$ ,
$2\cdot3^2=18$
and $10+4(2)=18$ , so neither exceeds the other yet.

At $n=3$ , A gives $54$ and B gives $22$ , so the first strict exceedance is $n=3$ .
The claim about eventual growth concerns sufficiently large inputs, not which value is larger initially.

Answer

$n=3$ ; A reports $54$ and B reports $22$ .

Key idea

Eventual exponential dominance permits a linear quantity to be larger at early inputs.
Problem 7: Mark complete
Problem 8 Two outputs multiplied

Let $F(x)=4b^x$ , where $b>0$ and $b\ne1$ . Express the product $F(s)F(t)$ in the form $cF(s+t)$ for a constant $c$ , and explain why $c$ is not $1$ .
1. Hint 1
  Adding exponents is what turns a product of two powers of one base into a single power.
2. Hint 2
  Expand both values, combine the powers of $b$ , then compare the result with $F(s+t)=4b^{s+t}$ .
Answer

$F(s)F(t)=4F(s+t)$ , so $c=4$ .

Full solution

Expanding both values gives
$F(s)F(t)=16b^sb^t$
The exponent rule combines the two powers:
$16b^sb^t=16b^{s+t}$

A single output is $F(s+t)=4b^{s+t}$ , so the product is
$F(s)F(t)=4F(s+t)$

The constant is not $1$ because the multiplier $4$ is used twice on the left and only once in $F(s+t)$ .

Answer

$F(s)F(t)=4F(s+t)$ , so $c=4$ .

Key idea

Multiplying two outputs of $a\cdot b^x$ adds the exponents but uses the multiplier twice, so the product is $a$ times the output at the summed input.
Problem 8: Mark complete
Problem 9 The three readings

The readings at inputs $0$ , $1$ , and $2$ are $2$ , $4$ , and $8$ . Eli says these three readings prove that the unknown rule is $f(x)=2\cdot2^x$ at every real input. Compare this proposed rule with $g(x)=x^2+x+2$ at those inputs and at $x=3$ , then decide whether Eli is right.
1. Hint 1
  A finite set of matching readings can belong to more than one rule.
2. Hint 2
  Test whether the polynomial matches the three records before comparing a new input.
Answer

Both give $2,4,8$ at $0,1,2$ ; at $3$ , $f(3)=16$ and $g(3)=14$ . Eli is wrong.

Full solution

The polynomial gives $g(0)=2$ , $g(1)=4$ , and $g(2)=8$ , exactly matching the exponential at all three recorded inputs.

At the new input,
$\begin{aligned}f(3)&=2\cdot2^3\\f(3)&=16\\g(3)&=3^2+3+2\\g(3)&=14\end{aligned}$

The records fit the exponential pattern, but they also fit a different rule that later disagrees.
They therefore do not prove the unknown rule at every input.

Answer

Both give $2,4,8$ at $0,1,2$ ; at $3$ , $f(3)=16$ and $g(3)=14$ . Eli is wrong.

Key idea

Constant ratios in a finite table support an exponential model without uniquely determining an unknown function everywhere.
Problem 9: Mark complete
Problem 10 The two bases

Let $f(x)=3^x$ and $g(x)=5^x$ . Bea says $g(x)>f(x)$ for every real $x\ne0$ since its base is larger. Is she right? Compare one positive input and one negative input to justify your decision.
1. Hint 1
  A negative exponent takes a reciprocal, which can reverse a comparison of positive values.
2. Hint 2
  The inputs $1$ and $-1$ make both comparisons direct.
Answer

No; at $1$ , $g(1)=5>3=f(1)$ ; at $-1$ , $g(-1)=\frac15<\frac13=f(-1)$ .

Full solution

At input $1$ , the outputs are $3$ and $5$ , so the larger base gives the larger output.
At input $-1$ ,
$\begin{aligned}f(-1)&=\frac13\\g(-1)&=\frac15\end{aligned}$

A fifth is smaller than a third, so
$g(-1)<f(-1).$
One negative input is enough to refute Bea's claim that $g(x)>f(x)$ at every nonzero input, even though the claim does hold for every positive input.

Answer

No; at $1$ , $g(1)=5>3=f(1)$ ; at $-1$ , $g(-1)=\frac15<\frac13=f(-1)$ .

Key idea

A larger exponential base gives larger positive-input powers, but taking reciprocals reverses the comparison at negative inputs.
Problem 10: Mark complete

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