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Exponential and Logarithmic Functions: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 An equality between three powers

Difficulty: 1 of 3 stars, Stretch

Find every real solution of $6^x+8^x=10^x$ . Then determine exactly when $6^x+8^x>10^x$ and when $6^x+8^x<10^x$ . Your justification must cover negative as well as positive exponents.
1. Hint 1
  Divide by the positive quantity $10^x$ . What happens to each resulting term when $x$ increases?
2. Hint 2
  Look for one exact solution using a familiar relation among $6,8,10$ , then use strict monotonicity to rule out another.
Answer

Equality holds only at $x=2$ . The left side is greater for $x<2$ and smaller for $x>2$ .

Full solution

Division by $10^x$ preserves the equation and both inequality directions because $10^x>0$ for every real $x$ .
Thus we compare $F(x)=(3/5)^x+(4/5)^x$ with $1$ .
Both bases lie strictly between $0$ and $1$ , so each summand strictly decreases as $x$ increases.
Their sum therefore strictly decreases on the entire real line.

At $x=2$ , the value is
$9/25+16/25=1.$
Consequently $F(x)>F(2)=1$ whenever $x<2$ , and $F(x)<F(2)=1$ whenever $x>2$ .
Multiplying by $10^x$ gives all the stated conclusions.
In particular, the same argument covers $x=0$ and all negative exponents; restricting a numerical search to positive integers would not establish completeness.

Answer

Equality holds only at $x=2$ . The left side is greater for $x<2$ and smaller for $x>2$ .

Key idea

Normalize an exponential equation so that every remaining term moves in the same direction.
Problem 1: Mark complete
Problem 2 Two matching marks

Difficulty: 1 of 3 stars, Stretch

Let $F(x)=\log_2(x-a)+b$ and $G(x)=\log_2(3x+6)$ , where $a,b$ are real. Both functions are defined at $x=0$ and $x=2$ , and $F(0)=G(0)$ and $F(2)=G(2)$ . Find $a,b$ and the common domain.

Prove that these two matching marks force $F(x)=G(x)$ everywhere in that domain. More generally, explain why any two functions of the form $\log_2(ux+v)+w$ , with $u>0$ , are either identical on their common domain or have at most one intersection.
1. Hint 1
  Absorb an additive constant into the argument of a logarithm.
2. Hint 2
  After exponentiating, agreement at a point becomes agreement between two affine expressions.
Answer

$a=-2$ , $b=\log_2 3$ , with domain $x>-2$ ; the functions agree throughout it. The stated general intersection claim is true.

Full solution

Because $2^b>0$ , we may rewrite $F(x)=\log_2(2^b(x-a))$ on its original domain.
The logarithm is one-to-one, so the two given equalities become $-a2^b=6$ and
$(2-a)2^b=12.$
Subtracting yields $2\cdot2^b=6$ , hence $2^b=3$ , $b=\log_2 3$ , and $a=-2$ .
Both original logarithm arguments are positive exactly when $x>-2$ .

For these values, $2^b(x-a)=3(x+2)=3x+6$ , so equality holds throughout the common domain.
For the general claim, rewrite each function as the logarithm of an affine expression by moving its additive constant inside.
An intersection then solves equality of two affine expressions.
If their coefficients agree, the functions agree wherever both are defined.
Otherwise their difference is a nonzero affine expression, which has at most one zero.
This also explains why two genuinely independent matching marks are enough here.

Answer

$a=-2$ , $b=\log_2 3$ , with domain $x>-2$ ; the functions agree throughout it. The stated general intersection claim is true.

Key idea

Use a one-to-one transformation to turn agreement of nonlinear graphs into a simpler identity.
Problem 2: Mark complete
Problem 3 An integer report from three scales

Difficulty: 1 of 3 stars, Stretch

Find every positive integer $N$ for which $\log_2N+\log_8N+\log_{32}N$ is an integer. In particular, find the least such $N>1$ . Prove that your list is exhaustive.
1. Hint 1
  Express all three logarithms in base $2$ .
2. Hint 2
  If the sum equals an integer $k$ , exponentiation gives an equality between an integer power of $N$ and a power of $2$ . Use prime factorization.
Answer

$N=2^{15m}$ for integers $m\ge0$ ; the least value greater than $1$ is $2^{15}=32768$ .

Full solution

Change of base gives the sum as
$(1+1/3+1/5)\log_2N=(23/15)\log_2N.$
Since $N$ is a positive integer, this sum is nonnegative.
If it equals an integer $k\ge0$ , then
$N^{23}=2^{15k}.$
Unique prime factorization forces $N$ itself to be a power of $2$ : write $N=2^j$ with integer $j\ge0$ .
The exponent equation is $23j=15k$ .

Because $23$ and $15$ are relatively prime, $15$ must divide $j$ .
Thus $j=15m$ for an integer $m\ge0$ , giving the proposed list.
Conversely, for $N=2^{15m}$ , the three logarithms are $15m,5m,3m$ , whose sum is the integer $23m$ .
This proves sufficiency as well as necessity.
The case $m=0$ gives $N=1$ ; the first larger case is $m=1$ .

Answer

$N=2^{15m}$ for integers $m\ge0$ ; the least value greater than $1$ is $2^{15}=32768$ .

Key idea

An integer condition on logarithms can become a divisibility condition on prime exponents.
Problem 3: Mark complete
Problem 4 A mixture that changes its apparent rate

Difficulty: 2 of 3 stars, Challenge

A sample contains two substances that decay independently and exponentially. Substance A has a half-life of $3$ hours; substance B has a half-life of $6$ hours. The total amount is initially $100$ units and is $35$ units after $6$ hours. Find the initial amount of each substance.

Let $M(t)$ be the total amount at time $t\ge0$ , and define $R(t)=M(t+6)/M(t)$ . Prove that $R(t)$ strictly increases with $t$ , even though both substances keep their original half-lives. Explain why one exponential function cannot describe the total at all times.
1. Hint 1
  Over $6$ hours, A is multiplied by $1/4$ and B by $1/2$ .
2. Hint 2
  After finding the initial amounts, write $q=2^{-t/6}$ and simplify $R(t)$ into a constant plus one positive fraction.
Answer

Initially A is $60$ units and B is $40$ units. $R(t)=1/4+10/(60\cdot2^{-t/6}+40)$ , which strictly increases; no single exponential describes $M$ .

Full solution

Write the initial amounts as $A,B$ .
The measurements give $A+B=100$ and
$A/4+B/2=35.$
Multiplying the second equation by $4$ and subtracting the first gives $B=40$ , so $A=60$ .
Thus
$M(t)=60\cdot2^{-t/3}+40\cdot2^{-t/6}.$

Put
$q=2^{-t/6}>0.$
Then $M(t)=60q^2+40q$ and
$M(t+6)=15q^2+20q.$
Canceling the positive factor $q$ gives
$R(t)=(15q+20)/(60q+40)=1/4+10/(60q+40).$
As $t$ increases, $q$ strictly decreases, so the positive denominator decreases and this final fraction strictly increases.
In particular, $1/4<R(t)<1/2$ for every finite $t\ge0$ .

For a single exponential $Ce^{kt}$ with $C>0$ , the ratio between readings $6$ hours apart is always $e^{6k}$ , independent of the starting time.
The increasing ratio here rules out such a model.
The slower-decaying component becomes a larger share of the mixture, so the total loses a smaller fraction during later intervals.

Answer

Initially A is $60$ units and B is $40$ units. $R(t)=1/4+10/(60\cdot2^{-t/6}+40)$ , which strictly increases; no single exponential describes $M$ .

Key idea

A sum of exponentials need not have a constant proportional rate; compare equal-time ratios.
Problem 4: Mark complete
Problem 5 An operating window

Difficulty: 2 of 3 stars, Challenge

For $t\ge0$ , a model assigns a total load $L(t)=2e^t+3e^{-2t}$ . Determine the complete time interval on which $L(t)\le5$ .

Also find the least possible load and the time when it occurs. Prove global minimality without using calculus.
1. Hint 1
  Put $u=e^t$ , remembering that $u\ge1$ . Multiply the load inequality by $u^2$ .
2. Hint 2
  For the minimum, let $a=\sqrt[3]{3}$ and try factoring $u^2(L-3a)$ .
Answer

$0\le t\le\ln((3+\sqrt{33})/4)$ . The minimum is $3\sqrt[3]{3}$ , attained only at $t=(\ln3)/3$ .

Full solution

Let $u=e^t\ge1$ .
The inequality becomes $2u+3/u^2\le5$ , equivalently
$(u-1)(2u^2-3u-3)\le0.$
The quadratic roots are $(3\pm\sqrt{33})/4$ .
The negative root is irrelevant, and the positive root $r=(3+\sqrt{33})/4$ exceeds $1$ .
On $u\ge1$ , the product is nonpositive exactly for
$1\le u\le r.$
Applying the increasing natural logarithm gives the reported interval, including both endpoints.

To locate the minimum without guessing from a graph, balance the two powers using
$a=\sqrt[3]{3}.$
Direct factorization gives
$L-3a=(u-a)^2(2u+a)/u^2.$
Every factor except the square is positive for $u\ge1$ , so $L\ge3a$ .
Equality holds precisely at $u=a$ , which is allowed because $a>1$ .
Since $e^t=a$ is equivalent to $t=(\ln3)/3$ , this proves both global minimality and uniqueness of the minimizing time.

Answer

$0\le t\le\ln((3+\sqrt{33})/4)$ . The minimum is $3\sqrt[3]{3}$ , attained only at $t=(\ln3)/3$ .

Key idea

After an exponential substitution, a well-chosen factorization can certify both a time window and a global bound.
Problem 5: Mark complete
Problem 6 A sum of two powers

Difficulty: 2 of 3 stars, Challenge

Find all triples of nonnegative integers $(a,b,c)$ with $a\le b$ such that $2^a+2^b=3^c$ . Prove that no larger exponents produce additional solutions.
1. Hint 1
  First use parity to determine the smaller exponent. Separate the few smallest values of the remaining exponent.
2. Hint 2
  For the remaining cases, compare powers of $3$ with multiples of $8$ . If $c$ is even, factor $3^c-1$ as a difference of squares.
Answer

The only triples are $(0,1,1)$ and $(0,3,2)$ .

Full solution

If $a=b$ , then $2^a+2^b=2^{a+1}$ is even, whereas $3^c$ is odd.
Thus $a<b$ .
Factoring gives
$2^a(1+2^{b-a})=3^c.$
The parenthesis is odd, so parity forces $a=0$ , leaving $1+2^b=3^c$ with $b\ge1$ .
For $b=1$ we obtain $c=1$ ; for $b=2$ the value $5$ is not a power of $3$ .

Now suppose $b\ge3$ .
The left side leaves remainder $1$ upon division by $8$ .
Powers of $3$ alternate between remainders $3$ and $1$ , since $3^2$ leaves remainder $1$ .
Consequently $c=2d$ for a positive integer $d$ .
The equation factors as
$(3^d-1)(3^d+1)=2^b.$
Both positive factors are even and must be powers of $2$ .
Write them as $2^u,2^v$ with $1\le u<v$ .
Their difference is $2$ , so
$2^u(2^{v-u}-1)=2.$
The second factor is odd, forcing $u=1$ and then $v=2$ .
Thus $3^d=3$ , giving $d=1$ , $c=2$ , and $b=3$ .

Direct substitution verifies $1+2=3$ and $1+8=9$ .
The parity and factorization arguments cover all possible larger exponents, so the list is exhaustive.

Answer

The only triples are $(0,1,1)$ and $(0,3,2)$ .

Key idea

For integer exponential equations, a small remainder pattern can expose a factorization that bounds every exponent.
Problem 6: Mark complete
Problem 7 A conclusion with a missing hypothesis

Difficulty: 2 of 3 stars, Challenge

For $x,y,z>1$ , suppose $\log_x y+\log_y z+\log_z x=3$ . Prove that $x=y=z$ .

A student claims that the same conclusion holds whenever $x,y,z$ are merely positive and different from $1$ . Decide whether this is true. If it is false, give an exact counterexample and identify the failed step in the original argument.
1. Hint 1
  The three logarithms have product $1$ . Under the first hypothesis they are also positive.
2. Hint 2
  For positive $A,B,C$ with $ABC=1$ , put $p=\sqrt[3]{A}$ , $q=\sqrt[3]{B}$ , $r=\sqrt[3]{C}$ and use the factorization of $p^3+q^3+r^3-3pqr$ . For a counterexample, try two equal negative logarithms.
Answer

Under $x,y,z>1$ , necessarily $x=y=z$ . With only positivity and bases different from $1$ , $(x,y,z)=(16,1/4,2)$ is a counterexample.

Full solution

Let $A=\log_x y$ , $B=\log_y z$ , and
$C=\log_z x.$
Change of base gives $ABC=1$ .
When $x,y,z>1$ , all three are positive.
For $p=\sqrt[3]{A}$ , $q=\sqrt[3]{B}$ , and $r=\sqrt[3]{C}$ , we have $pqr=1$ and

$A+B+C-3=\frac{p+q+r}{2}\bigl((p-q)^2+(q-r)^2+(r-p)^2\bigr).$

The right side is nonnegative and vanishes only when $p=q=r$ .
Since the given sum is $3$ , equality holds.
The product condition then gives $A=B=C=1$ , hence $y=x$ , $z=y$ , and $x=z$ .
Conversely, every triple with $x=y=z>1$ satisfies the equation.

For $(16,1/4,2)$ , the three logarithms are $-1/2,-1/2,4$ , whose sum is $3$ and product is $1$ , but the bases are not equal.
The product condition survives; positivity of the logarithms does not.
The factorization still holds, but its prefactor need not be positive when signed cube roots are used, so it no longer forces all three squares to vanish.

Answer

Under $x,y,z>1$ , necessarily $x=y=z$ . With only positivity and bases different from $1$ , $(x,y,z)=(16,1/4,2)$ is a counterexample.

Key idea

A proof that uses positivity must check the signs of transformed quantities, not just the original variables.
Problem 7: Mark complete
Problem 8 Equal powers with unequal bases

Difficulty: 3 of 3 stars, Deep challenge

Find a parametrization of every pair of positive real numbers $x<y$ satisfying $x^y=y^x$ . Your parameter should be the ratio $r=y/x>1$ , and you must prove the converse.

Among these pairs, find all those for which both $x$ and $y$ are integers. Prove completeness without calculus.
1. Hint 1
  Take natural logarithms, substitute $y=rx$ , and divide only by quantities known to be positive.
2. Hint 2
  For integer $x<y$ , write $x=gu$ , $y=gv$ with $g$ their greatest common divisor. The equation gives $g^{v-u}u^v=v^u$ . What does a prime divisor of $u$ imply? Then compare $2^{m-1}$ with $m$ for integers $m\ge3$ .
Answer

$x=r^{1/(r-1)}$ , $y=r^{r/(r-1)}$ for arbitrary $r>1$ . The only integer pair with $x<y$ is $(2,4)$ .

Full solution

Taking logarithms of the original equality gives
$y\ln x=x\ln y.$
Since $x>0$ and $y=rx$ , division by $x$ gives
$r\ln x=\ln r+\ln x.$
Therefore $(r-1)\ln x=\ln r$ , so $x=r^{1/(r-1)}$ and
$y=rx=r^{r/(r-1)}.$
Conversely, for every $r>1$ , these positive numbers have ratio $r$ and satisfy the derived logarithmic equality.
Exponentiating recovers $x^y=y^x$ , proving completeness.

For integer $x<y$ , let $g$ be their greatest common divisor and write $x=gu$ , $y=gv$ , where $u,v$ are relatively prime positive integers and $u<v$ .
Taking the positive $g$ th root of the power equation and canceling $g^u$ gives
$g^{v-u}u^v=v^u.$
If a prime divided $u$ , it would divide the left side but not $v^u$ , a contradiction.
Hence $u=1$ , so the ratio $y/x=v$ must be an integer.

Write this integer ratio as $r=m\ge2$ .
The formula gives $x>1$ , hence $x\ge2$ , and $x^{m-1}=m$ .
If $m\ge3$ , then $2^{m-1}>m$ : this holds at $m=3$ , and multiplying by $2$ gives $2^m>2m\ge m+1$ , establishing the next case.
Thus $x^{m-1}\ge2^{m-1}>m$ , a contradiction.
The remaining value $m=2$ gives $x=2$ and $y=4$ , which satisfies all conditions.
The order $x<y$ excludes reversing the pair.

Answer

$x=r^{1/(r-1)}$ , $y=r^{r/(r-1)}$ for arbitrary $r>1$ . The only integer pair with $x<y$ is $(2,4)$ .

Key idea

A ratio parameter can turn two variables occurring in exponents into a complete one-parameter family.
Problem 8: Mark complete
Problem 9 Dividing time between two stages

Difficulty: 3 of 3 stars, Deep challenge

A process has a fixed total duration $T>0$ . If the first stage lasts $s$ time units, it produces $e^s$ units of intermediate material. During the remaining $T-s$ units, the second stage converts the fraction $1-e^{-(T-s)}$ of that material into finished output. Assume $0\le s\le T$ and that these formulas are exact.

For every $T>0$ , determine the greatest finished output and every allocation $s$ attaining it. Explain any change in the form of the answer.

What is the least total duration that permits at least $9/2$ units of finished output, and how must that duration be divided?

Builds on Completing the Square
1. Hint 1
  Put $u=e^s$ . Its allowed interval depends on $T$ , and the output becomes a quadratic in $u$ .
2. Hint 2
  Complete the square. Check whether the vertex lies inside $1\le u\le e^T$ before accepting it as an optimizer.
Answer

For $0<T\le\ln2$ , the maximum is $1-e^{-T}$ at $s=0$ . For $T\ge\ln2$ , it is $e^T/4$ at $s=T-\ln2$ ; the formulas agree at $T=\ln2$ . The least duration for $9/2$ units is $T=\ln18$ , split as $s=\ln9$ and $T-s=\ln2$ .

Full solution

The output is
$Y=e^s(1-e^{-(T-s)}).$
Set $u=e^s$ , so
$1\le u\le e^T.$
Then
$Y=u-e^{-T}u^2=e^T/4-e^{-T}(u-e^T/2)^2.$
Maximizing output is therefore the same as making $u$ as close as possible to $e^T/2$ within the allowed interval.

If $T\le\ln2$ , the vertex $e^T/2$ lies at or below the left endpoint $1$ .
The unique optimum is $u=1$ , or $s=0$ , with output $1-e^{-T}$ .
If $T\ge\ln2$ , the vertex is allowed and gives the unique optimum $u=e^T/2$ , or $s=T-\ln2$ , with output $e^T/4$ .
At the boundary both descriptions give $s=0$ and output $1/2$ .
The change comes from the constraint $s\ge0$ , not from a failure of the quadratic formula.

The first range of durations gives output at most $1/2$ , so it cannot reach $9/2$ .
In the second range, reaching that target is possible exactly when $e^T/4\ge9/2$ , or $T\ge\ln18$ .
At the least duration equality forces the unique maximizing allocation: $s=\ln18-\ln2=\ln9$ , leaving $\ln2$ for the second stage.

Answer

For $0<T\le\ln2$ , the maximum is $1-e^{-T}$ at $s=0$ . For $T\ge\ln2$ , it is $e^T/4$ at $s=T-\ln2$ ; the formulas agree at $T=\ln2$ . The least duration for $9/2$ units is $T=\ln18$ , split as $s=\ln9$ and $T-s=\ln2$ .

Key idea

An unconstrained optimum is only a candidate; the allowed interval can change both the optimum and its interpretation.
Problem 9: Mark complete
Problem 10 Choosing a logarithmic scale

Difficulty: 3 of 3 stars, Deep challenge

A designer chooses a base $b>1$ and places a positive number $v$ at coordinate $\log_b v$ . The desired coordinates for $v=2,3,10$ are $1,2,3$ , respectively.

Determine the unique base that minimizes the largest coordinate error
$E(b)=\max\{|\log_b2-1|,\ |\log_b3-2|,\ |\log_b10-3|\}.$
Find the minimum error exactly. Prove that no other base does as well, and verify all three errors at your proposed base.

Builds on Linear Inequalities
1. Hint 1
  Use $u=1/\ln b>0$ so that each logarithm becomes a constant times $u$ .
2. Hint 2
  If the error is at most $E$ , the targets at $3$ and $10$ impose $u\ln3\ge2-E$ and $u\ln10\le3+E$ . Combine them to obtain a necessary lower bound for $E$ .
Answer

The unique optimum is $b=30^{1/5}$ . The minimum error is $\ln(100/27)/\ln30$ .

Full solution

Put $c=\ln3$ , $d=\ln10$ , and
$u=1/\ln b>0.$
This substitution is one-to-one between bases $b>1$ and numbers $u>0$ .
If all three errors are at most $E$ , then $cu\ge2-E$ and $du\le3+E$ .
Multiplying by positive constants and eliminating $u$ gives
$d(2-E)\le c(3+E).$
Therefore every base satisfies
$E\ge E_*=(2d-3c)/(c+d)=\ln(100/27)/\ln30>0.$

To attain this bound, both inequalities must be equalities.
They give $u_*=5/(c+d)=5/\ln30$ , hence
$b=e^{1/u_*}=30^{1/5}.$
At this base, the errors at $3$ and $10$ are exactly $E_*$ , one below its target and one above.
The remaining error is
$|5\ln2/\ln30-1|=\ln(16/15)/\ln30.$
It is smaller than $E_*$ because
$1<16/15<100/27.$
Thus the proposed base really attains the lower bound for all three marks.

Finally, if another base attained $E_*$ , the lower and upper bounds on $u$ obtained from the marks at $3$ and $10$ would coincide.
This forces $u=u_*$ and therefore the same base.
The third mark is a necessary feasibility check even though it does not determine the optimum.

Answer

The unique optimum is $b=30^{1/5}$ . The minimum error is $\ln(100/27)/\ln30$ .

Key idea

To prove a minimax optimum, derive a lower bound from conflicting requirements and then verify every remaining requirement.
Problem 10: Mark complete