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Polynomials, Exponentials, and Logarithms: 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 When every odd power disappears

Difficulty: 1 of 3 stars, Stretch

A polynomial is called even when replacing $x$ by $-x$ leaves it unchanged. Find all real numbers $a,b$ for which $(x^2+ax+2)(x^2+bx+8)$ is even.

Now replace $2,8$ by real constants $c,d$ . Classify all pairs $(a,b)$ for which $(x^2+ax+c)(x^2+bx+d)$ is even, stating how the answer depends on $c,d$ .
1. Hint 1
  Which coefficients must vanish in an even polynomial?
2. Hint 2
  Compute only the coefficients of $x^3$ and $x$ , then substitute one resulting equation into the other.
Answer

For $c=2,d=8$ , only $(a,b)=(0,0)$ . Generally, if $c\ne d$ , only $(0,0)$ ; if $c=d$ , every pair with $b=-a$ .

Full solution

An even polynomial has zero coefficients on its odd powers: subtraction of its value at $-x$ from its value at $x$ leaves exactly twice those terms.
In this product the only possible odd powers are $x^3$ and $x$ .
Their coefficients are $a+b$ and $ad+bc$ , respectively.

Thus evenness is equivalent to the two equations $a+b=0$ and $ad+bc=0$ .
Substituting $b=-a$ into the second gives $a(d-c)=0$ .
If $c\ne d$ , this forces $a=0$ and then $b=0$ .
In particular this covers $c=2,d=8$ .

If $c=d$ , the second equation follows automatically from $b=-a$ , so $a$ can be any real number.
Conversely, every pair in the stated lists makes both odd coefficients zero.
This verifies sufficiency as well as necessity, including zero or negative choices of $c,d$ .

Answer

For $c=2,d=8$ , only $(a,b)=(0,0)$ . Generally, if $c\ne d$ , only $(0,0)$ ; if $c=d$ , every pair with $b=-a$ .

Key idea

Compare only the coefficients that control the requested property; retain parameter values that make an equation redundant.
Problem 1: Mark complete
Problem 2 Growth behind a constant background

Difficulty: 1 of 3 stars, Stretch

A sensor reading after $n$ hours is $F(n)=A\,2^n+B$ , where $A>0$ , $B$ is real, and $n$ is a nonnegative integer. The readings are $F(2)=17$ and $F(5)=73$ .

Find the initial reading, the least integer $n$ for which $F(n)>1000$ , and the amount by which consecutive readings increase. Explain why the readings themselves do not double each hour.

Builds on Exponential Functions
1. Hint 1
  Subtract the two observed readings to eliminate the background.
2. Hint 2
  After finding $A,B$ , compare the powers $2^8$ and $2^9$ with the threshold. Compute $F(n+1)-F(n)$ .
Answer

$A=2$ , $B=9$ , $F(0)=11$ ; the least $n$ is $9$ . The increase from hour $n$ to hour $n+1$ is $2^{n+1}$ .

Full solution

The observations give $4A+B=17$ and $32A+B=73$ .
Subtraction yields $28A=56$ , so $A=2$ and $B=9$ .
Therefore $F(n)=2^{n+1}+9$ and the initial reading is $F(0)=11$ .

The inequality $F(n)>1000$ is equivalent to
$2^{n+1}>991.$
Since $2^9=512<991<1024=2^{10}$ , hour $9$ is the first integer hour that works.
Indeed, $F(8)=521$ and $F(9)=1033$ .
The expression increases with $n$ , which rules out every earlier hour.

Consecutive differences satisfy
$F(n+1)-F(n)=2^{n+2}-2^{n+1}=2^{n+1}.$
These differences double.
The readings obey $F(n+1)=2F(n)-9$ , rather than $F(n+1)=2F(n)$ : doubling a reading would incorrectly double the constant background too.

Answer

$A=2$ , $B=9$ , $F(0)=11$ ; the least $n$ is $9$ . The increase from hour $n$ to hour $n+1$ is $2^{n+1}$ .

Key idea

A constant offset can hide exponential growth; differences remove the offset.
Problem 2: Mark complete
Problem 3 Two exponentials moving in opposite directions

Difficulty: 1 of 3 stars, Stretch

Find all real solutions of $3^x+3^{4-x}=30$ . Then find the least possible value of $3^x+3^{4-x}$ over all real $x$ , and prove that your value is the global minimum.

Builds on Factoring Quadratics
1. Hint 1
  The two positive terms have a constant product.
2. Hint 2
  Put $t=3^x$ . For the minimum, rewrite $t+81/t-18$ as a square divided by $t$ .
Answer

The solutions are $x=1,3$ . The minimum is $18$ , attained only at $x=2$ .

Full solution

Put $t=3^x$ , so $t>0$ and
$3^{4-x}=81/t.$
The equation becomes $t+81/t=30$ .
Multiplying by the positive, nonzero $t$ gives $t^2-30t+81=0$ , or
$(t-3)(t-27)=0.$
Thus $3^x=3$ or $27$ , yielding $x=1$ or $3$ .
Both give $3+27=30$ in the original equation.

For every $t>0$ , the identity $t+81/t-18=(t-9)^2/t$ has a nonnegative right side.
Hence the expression is at least $18$ , with equality exactly when $t=9$ .
This corresponds to $x=2$ .
The argument covers all positive $t$ , and every real $x$ produces such a $t$ , so this is a global bound rather than a comparison of selected inputs.

Answer

The solutions are $x=1,3$ . The minimum is $18$ , attained only at $x=2$ .

Key idea

Reciprocal exponential terms often become one positive variable and a useful square.
Problem 3: Mark complete
Problem 4 A product with no missing powers

Difficulty: 2 of 3 stars, Challenge

Let $P(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$ . Prove that every coefficient from $x^0$ through $x^{15}$ in $P(x)$ is $1$ , without multiplying all four factors term by term.

Use that structure to find the coefficients of $x^{15}$ and $x^{20}$ in $P(x)^2$ .

Builds on Difference of Squares
1. Hint 1
  Consider multiplying the product by $1-x$ .
2. Hint 2
  After identifying $P(x)$ , a term $x^m$ in its square comes from ordered exponent pairs $(i,j)$ with $i+j=m$ and $0\le i,j\le15$ .
Answer

$P(x)=1+x+\cdots+x^{15}$ . The requested coefficients in $P(x)^2$ are $16$ and $11$ , respectively.

Full solution

Repeatedly use the difference of squares to obtain
$(1-x)P(x)=1-x^{16}.$
Also, direct cancellation gives
$(1-x)(1+x+\cdots+x^{15})=1-x^{16}.$
Subtract the two identities.
Their difference says that $(1-x)$ times the difference of the two polynomials is zero for every $x$ .

The difference polynomial must be zero: if it had a highest nonzero coefficient, multiplying by $1-x$ would produce a nonzero coefficient one degree higher.
Consequently
$P(x)=1+x+\cdots+x^{15}.$
This polynomial argument also includes $x=1$ , where numerical division by $1-x$ would be invalid.

In the square, each ordered pair $(i,j)$ of exponents contributes one $x^{i+j}$ .
For $i+j=15$ , all $i=0,1,\ldots,15$ work, giving $16$ .
For $i+j=20$ , the requirements $0\le20-i\le15$ and $0\le i\le15$ give $5\le i\le15$ , giving $11$ .
The pairs are ordered because the two chosen terms come from two separate copies of $P$ .

Answer

$P(x)=1+x+\cdots+x^{15}$ . The requested coefficients in $P(x)^2$ are $16$ and $11$ , respectively.

Key idea

An auxiliary factor can reveal a whole polynomial; coefficients of a product count ways to combine exponents.
Problem 4: Mark complete
Problem 5 Interest with an invisible yearly fee

Difficulty: 2 of 3 stars, Challenge

An account earns a fixed positive annual interest rate. At the end of each year, interest is added first and then the same fixed fee is subtracted. There are no other deposits or withdrawals. Its balances after years $1,2,3$ are $1020$ , $1140$ , and $1284$ dollars.

Find the annual interest rate, fee, and initial balance. Then determine the least whole number of years after opening when the balance reaches at least twice its initial value. Give exact comparisons, without a logarithm approximation.

Builds on Compound Interest, Exponential Functions
1. Hint 1
  If the interest multiplier is $q$ , write $B_{n+1}=qB_n-f$ . Subtract two consecutive equations.
2. Hint 2
  Find a constant $L$ satisfying $L=qL-f$ . The amounts $B_n-L$ then multiply by $q$ each year.
Answer

The rate is $20\%$ , the fee is $84$ dollars, and the initial balance is $920$ dollars. The balance first reaches twice its initial value after $6$ years.

Full solution

Write $q=1+r>1$ for the interest multiplier and $f$ for the fee.
The reports give $1140=1020q-f$ and
$1284=1140q-f.$
Subtraction yields $144=120q$ , so $q=6/5$ and $r=1/5$ .
Substitution gives $f=84$ .
Finally, $1020=(6/5)B_0-84$ gives $B_0=920$ .

The balance $420$ would remain unchanged, since
$(6/5)420-84=420.$
Subtracting $420$ from the yearly rule therefore gives
$B_{n+1}-420=(6/5)(B_n-420).$
Starting with $B_0-420=500$ and applying this relation repeatedly yields
$B_n=420+500(6/5)^n.$

Twice the initial balance is $1840$ , so we need
$(6/5)^n\ge71/25.$
At $n=5$ , compare $7776/3125$ with $8875/3125$ ; the inequality fails.
At $n=6$ , compare $46656/15625$ with $44375/15625$ ; it holds.
Since $6/5>1$ , these powers increase strictly, proving that $6$ is the first year.

Answer

The rate is $20\%$ , the fee is $84$ dollars, and the initial balance is $920$ dollars. The balance first reaches twice its initial value after $6$ years.

Key idea

Subtracting a balance that would stay fixed can turn interest with a fee into pure exponential growth.
Problem 5: Mark complete
Problem 6 Recovering a cubic from two reports

Difficulty: 2 of 3 stars, Challenge

A polynomial has the form $P(x)=(x-a)(x-b)(x-c)$ , where $a,b,c$ are integers greater than $1$ , with repetition allowed. You are told that $P(1)=-12$ and $P(-1)=-90$ . Find $P(x)$ in expanded form, and prove that it is uniquely determined.
1. Hint 1
  What products of shifted roots do the evaluations at $1$ and $-1$ determine?
2. Hint 2
  Put $u=a-1$ , $v=b-1$ , $w=c-1$ . List positive integer triples with $uvw=12$ , ignoring their order, and check $(u+2)(v+2)(w+2)$ .
Answer

$P(x)=(x-2)(x-4)(x-5)=x^3-11x^2+38x-40$ .

Full solution

Using the given form $P(x)=(x-a)(x-b)(x-c)$ , the two evaluations yield $(a-1)(b-1)(c-1)=12$ and
$(a+1)(b+1)(c+1)=90.$

Set $u=a-1$ , $v=b-1$ , $w=c-1$ .
These are positive integers with $uvw=12$ .
We may arrange $u\le v\le w$ because permuting roots does not change $P$ .
Since $u^3\le12$ , either $u=1$ or $u=2$ .
If $u=1$ , the possible factor pairs for $vw=12$ are $(1,12),(2,6),(3,4)$ .
If $u=2$ , the constraints $vw=6$ and $2\le v\le w$ force
$(v,w)=(2,3).$

The four triples therefore give $(u+2)(v+2)(w+2)$ equal to $126,96,90,80$ , respectively.
Only $(u,v,w)=(1,3,4)$ works, so the roots are $2,4,5$ .
Expanding their monic product gives the stated polynomial, whose evaluations are indeed $-12$ and $-90$ .
The bounded factor list proves uniqueness.

Answer

$P(x)=(x-2)(x-4)(x-5)=x^3-11x^2+38x-40$ .

Key idea

Integer roots turn polynomial evaluations into a finite factor problem; a bound makes the list exhaustive.
Problem 6: Mark complete
Problem 7 When doubling overtakes a cube

Difficulty: 2 of 3 stars, Challenge

Find all nonnegative integers $n$ such that $2^n\ge n^3$ . Your argument must establish what happens for every larger integer after the last change, rather than relying on a long numerical table.

Builds on Exponential Functions
1. Hint 1
  Check small values to locate a possible final crossover.
2. Hint 2
  For $n\ge4$ , compare $(n+1)^3/n^3$ with $2$ . If doubling has already caught up, can the cube regain the lead?
Answer

$n=0$ , $n=1$ , and every integer $n\ge10$ .

Full solution

The inequality holds at $n=0,1$ , since $1\ge0$ and $2\ge1$ .
For $n=2,3,\ldots,9$ , the respective powers of two are $4,8,16,32,64,128,256,512$ , while the cubes are $8,27,64,125,216,343,512,729$ .
Each comparison fails.
At $n=10$ , it holds because $1024>1000$ .

It remains to prove persistence.
For every integer $n\ge4$ , $(n+1)/n\le5/4$ , so
$(n+1)^3/n^3\le125/64<2.$
If $2^n\ge n^3$ at such an $n$ , then
$2^{n+1}=2\cdot2^n\ge2n^3>(n+1)^3.$
Thus a successful value at or beyond $4$ forces success at the next integer.

Applying this implication successively from $10$ proves success at $11$ , then $12$ , and at every later integer.
Together with the complete small-value checks, it proves the classification.
The comparison concerns growth factors, which is why it controls an unlimited tail with one inequality.

Answer

$n=0$ , $n=1$ , and every integer $n\ge10$ .

Key idea

To prove that one growing expression stays ahead, compare the factors by which the expressions grow in one step.
Problem 7: Mark complete
Problem 8 A quartic whose roots change in groups

Difficulty: 3 of 3 stars, Deep challenge

For every real parameter $a$ , determine all distinct real roots of
$x^4-ax^3+(2a-2)x^2-ax+1=0.$
Give a complete classification of the number of distinct real roots, paying particular attention to parameter values at which roots merge.
1. Hint 1
  The coefficients read the same from either end. Since $x=0$ cannot be a root, divide by $x^2$ .
2. Hint 2
  Put $t=x+1/x$ . Factor the resulting quadratic in $t$ , then determine which real $t$ can arise from a real nonzero $x$ .
Answer

$x=1$ is always a root. For $a\le0$ or $a\ge4$ , also use $x=(a-2\pm\sqrt{a(a-4)})/2$ , counting duplicates only once. The counts are $3$ if $a<0$ , $2$ if $a=0$ , $1$ if $0<a\le4$ , and $3$ if $a>4$ .

Full solution

The constant term rules out $x=0$ .
After division by $x^2$ , use
$x^2+x^{-2}=(x+x^{-1})^2-2.$
With $t=x+x^{-1}$ the equation becomes
$t^2-at+2a-4=(t-2)(t-(a-2))=0.$

The branch $t=2$ gives $x^2-2x+1=0$ , hence $x=1$ .
The other branch gives $x^2-(a-2)x+1=0$ , with discriminant
$(a-2)^2-4=a(a-4).$
It has real roots exactly when $a\le0$ or $a\ge4$ , and then the quadratic formula gives the stated pair.
Their product is $1$ , so neither violates the original nonzero restriction.

If $a<0$ , the second branch gives two distinct negative roots, separate from $1$ , for a total of three.
At $a=0$ they merge to $-1$ , leaving two distinct roots.
If $0<a<4$ , neither is real, leaving only $1$ .
At $a=4$ , both merge to $1$ , again giving one.
If $a>4$ , they are distinct positive roots; neither equals $1$ , since substituting $1$ into their quadratic would require $a=4$ .
There are therefore three.

Every transformation was reversible for $x\ne0$ , so the classification contains all real roots and no extraneous ones.

Answer

$x=1$ is always a root. For $a\le0$ or $a\ge4$ , also use $x=(a-2\pm\sqrt{a(a-4)})/2$ , counting duplicates only once. The counts are $3$ if $a<0$ , $2$ if $a=0$ , $1$ if $0<a\le4$ , and $3$ if $a>4$ .

Key idea

Symmetry can reduce a quartic to two quadratic branches; count distinct roots only after checking branch overlap.
Problem 8: Mark complete
Problem 9 A moving logarithm base

Difficulty: 3 of 3 stars, Deep challenge

For each real parameter $a$ , determine all real solutions of $\log_{x-1}(x+a)=2$ . Classify the number of solutions, including every exceptional value of $a$ . Use real logarithms only.
1. Hint 1
  Write down the restrictions on the logarithm base before converting to an exponential equation.
2. Hint 2
  Put $y=x-1$ . Solve $y^2-y-(a+1)=0$ , but retain only $y>0$ with $y\ne1$ .
Answer

Candidate solutions are $x=(3\pm\sqrt{4a+5})/2$ , subject to $x>1$ and $x\ne2$ . There are no solutions if $a<-5/4$ or $a=-1$ ; one if $a=-5/4$ or $a>-1$ ; two if $-5/4<a<-1$ .

Full solution

A real logarithm requires $x-1>0$ , $x-1\ne1$ , and $x+a>0$ .
Put $y=x-1$ , so the base conditions become $y>0$ and $y\ne1$ .
The equation is equivalent to $x+a=y^2$ , or
$y^2-y-(a+1)=0.$
Any root satisfying the base conditions automatically has the positive argument $y^2$ , so no further argument restriction remains.

The quadratic gives
$y=(1\pm\sqrt{4a+5})/2.$
If $a<-5/4$ , these are not real.
At $a=-5/4$ , the single root $y=1/2$ is allowed and gives $x=3/2$ .
If $-5/4<a<-1$ , then $0<\sqrt{4a+5}<1$ , so both roots lie strictly between $0$ and $1$ ; both are allowed.

At $a=-1$ , the roots are $y=0,1$ , and both are forbidden bases.
If $a>-1$ , then the square root is greater than $1$ : the minus root is negative and the plus root is greater than $1$ .
Exactly the plus root is allowed.
Adding $1$ to each retained $y$ gives the candidate formula for $x$ .
All retained roots satisfy the original logarithm, since its argument is the square of its valid base.

Answer

Candidate solutions are $x=(3\pm\sqrt{4a+5})/2$ , subject to $x>1$ and $x\ne2$ . There are no solutions if $a<-5/4$ or $a=-1$ ; one if $a=-5/4$ or $a>-1$ ; two if $-5/4<a<-1$ .

Key idea

A logarithmic equation can lose solutions at forbidden bases even when its quadratic discriminant is positive.
Problem 9: Mark complete
Problem 10 A gap in the possible logarithm sums

Difficulty: 3 of 3 stars, Deep challenge

Let $x,y$ range over all positive real numbers with $xy=16$ and $x\ne1$ , $y\ne1$ . Determine the complete set of possible values of
$E=\log_x16+\log_y16.$
Prove both that every reported value is attainable and that no other value is. Also identify all pairs $(x,y)$ for which $E=4$ .
1. Hint 1
  Use $u=\log_2x$ and $v=\log_2y$ . What do the product condition and the two logarithms become?
2. Hint 2
  You will have $u+v=4$ and $E=16/(uv)$ . For a proposed value $E=t$ , ask whether a quadratic can have real nonzero roots $u,v$ with that sum and product.
Answer

The range is $(-\infty,0)\cup[4,\infty)$ . Equality $E=4$ occurs only at $(x,y)=(4,4)$ .

Full solution

Let $u=\log_2x$ and $v=\log_2y$ .
The domain restrictions give real nonzero $u,v$ , and $xy=16$ gives $u+v=4$ .
Since $16=2^4$ , changing the bases gives
$E=4/u+4/v=16/(uv).$

The square $(u-v)^2\ge0$ implies $uv\le4$ .
If $uv>0$ , then $E\ge4$ .
If $uv<0$ , then $E<0$ .
The product cannot be zero, so $E$ cannot be zero or fall between zero and four.
Equality at $4$ forces $uv=4$ and $u=v=2$ , giving $x=y=4$ .

To prove attainability, take any proposed real value $t<0$ or $t\ge4$ .
We seek roots $u,v$ of
$z^2-4z+16/t=0.$
Its discriminant is $16-64/t$ , which is positive for $t<0$ and nonnegative for $t\ge4$ .
Thus real roots exist.
Their product is $16/t\ne0$ , so neither is zero.

Define $x=2^u$ and $y=2^v$ .
They are positive and not $1$ , their product is $2^{u+v}=16$ , and their expression equals $16/(uv)=t$ .
This constructs an allowed pair for every value in the claimed range, completing both directions.

Answer

The range is $(-\infty,0)\cup[4,\infty)$ . Equality $E=4$ occurs only at $(x,y)=(4,4)$ .

Key idea

To prove an exact range, derive restrictions and then construct inputs for every surviving output.
Problem 10: Mark complete