Exponential and logarithmic functions · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

06 Exponential and logarithmic functions · g10-06-01 · 5–10 min · TEACHER · ANSWERS

A negative exponent is a reciprocal

Does 2⁻³ mean a negative number or a reciprocal?

Predict: Continue the sequence 2³,2²,2¹,2⁰ one step at a time to exponent −3.

Compare	Computed observation
Reciprocal 2⁻³ Base a = 2; Integer exponent n = -3	2^-3 = 0.125. A negative exponent takes a reciprocal.
Odd negative exponent Base a = -2; Integer exponent n = -3	-2^-3 = -0.125. A negative exponent takes a reciprocal.
Zero base forbidden Base a = 0; Integer exponent n = -1	Zero to a nonpositive exponent is undefined here.
Zero exponent Base a = 2; Integer exponent n = 0	2^0 = 1. A negative exponent takes a reciprocal.

Conclusion: Reducing the exponent by one divides by the nonzero base. Thus a⁰=1 and a⁻ⁿ=1/aⁿ for a≠0. The minus sign in the exponent does not negate the value.

Conditions: The exponent here is an integer. Zero to a negative exponent and 0⁰ are not defined in this task.

Justification: For a≠0, preserving aⁿa⁻ⁿ=a⁰=1 forces a⁻ⁿ=1/aⁿ. This algebra, rather than a few sequence terms, establishes the identity.

Ask: Ask students to name the repeated operation when the exponent decreases.

Watch for: a⁻ⁿ=−aⁿ.

Extension: For a=−2, compare odd and even negative exponents.

Exit: Find 5⁻².

1/25.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-02 · 5–10 min · TEACHER · ANSWERS

A fractional exponent names a root

What does 16^(3/4) ask you to do?

Predict: Take a fourth root and cube it; predict whether reversing those two operations changes the result.

Compare	Computed observation
Fourth root cubed Positive base a = 16; Integer numerator p = 3; Positive denominator q = 4	The positive 4th root is 2; raising it to 3 gives 8.
Negative numerator Positive base a = 16; Integer numerator p = -3; Positive denominator q = 4	The positive 4th root is 2; raising it to -3 gives 0.125.
Unit base Positive base a = 1; Integer numerator p = 3; Positive denominator q = 4	The positive 4th root is 1; raising it to 3 gives 1.
Nonpositive base rejected Positive base a = -1; Integer numerator p = 1; Positive denominator q = 2	This rational-power model requires a>0, integer p and positive integer q.

Conclusion: For positive a, a^(p/q)=(q√a)ᵖ with integer p and positive integer q. Here 16^(3/4)=2³=8. Negative p then takes a reciprocal.

Conditions: Use a>0 so all rational exponents have one consistent real definition. This activity rejects a≤0 rather than extending selected odd-root cases.

Justification: The positive qth root is the unique positive r with r^q=a. Define a^(p/q)=rᵖ; reducing p/q preserves the same positive value.

Ask: Label numerator “power” and denominator “root”; keep the positive-base condition beside the rule.

Watch for: a^(p/q)=(aᵖ)/q.

Extension: Explain why 16^(−3/4)=1/8.

Exit: Evaluate 27^(2/3).

9, since the cube root of 27 is 3.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-03 · 5–10 min · TEACHER · ANSWERS

Equal additions or equal ratios?

Does adding 20 each round match growing by 20% each round?

Predict: Starting at 100, predict both quantities after two rounds before reading the graph.

Compare	Computed observation
Two growth rounds Multiplier r = 1.2; Number of rounds n = 2	Round 2: fixed increment 140, fixed ratio 144.
Three decay rounds Multiplier r = 0.8; Number of rounds n = 3	Round 3: fixed increment 40, fixed ratio 51.2.
No change ratio 1 Multiplier r = 1; Number of rounds n = 3	Round 3: fixed increment 100, fixed ratio 100.
Initial round zero Multiplier r = 1.2; Number of rounds n = 0	Round 0: fixed increment 100, fixed ratio 100.

Conclusion: Fixed increments give L(n)=100+20n. Fixed ratios give E(n)=100×1.2ⁿ. They agree after one round but differ after two: 140 and 144.

Conditions: Rounds n are nonnegative integers. A constant ratio is an assumption. A negative linear-model prediction is not a meaningful remaining quantity.

Justification: The recurrence E(n+1)=rE(n), E(0)=100 gives E(n)=100rⁿ by induction. L(n+1)−L(n)=100(r−1) is constant, while exponential differences depend on n.

Ask: Ask for the next increment and next ratio in each model, not just the final values.

Watch for: A fixed percentage means the same absolute change each round.

Extension: Set multiplier 0.8 and compare repeated 20% removal with subtracting 20 each round.

Exit: After three rounds retaining 80% of 100, what remains?

100×0.8³=51.2.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-04 · 5–10 min · TEACHER · ANSWERS

Crossing the base 1 reverses growth

What changes in aˣ when a passes through 1?

Predict: Compare outputs at x=−1,0,1 for bases 0.5,1,2.

Compare	Computed observation
Growth base 2 Base a = 2; Input x = 1	a=2: increasing; a^1=2.
Decay base 0.5 Base a = 0.5; Input x = 1	a=0.5: decreasing; a^1=0.5.
Constant boundary 1 Base a = 1; Input x = 1	a=1: constant; a^1=1. Base 1 is the constant boundary, excluded from this course’s exponential family.
Nonpositive base Base a = 0; Input x = 1	A real exponential model requires a positive base.

Conclusion: For a>1, aˣ is increasing; for 0<a<1 it is decreasing. Every positive base gives a⁰=1. At a=1 the rule is constant, not an exponential function in this course.

Conditions: The real exponential family requires a>0 and a≠1. Its domain is all real numbers and its range is (0,∞).

Justification: For x₂>x₁, a^x₂/a^x₁=a^(x₂−x₁). This ratio exceeds 1 if a>1 and is below 1 if 0<a<1.

Ask: Keep the point (0,1) fixed while comparing a positive and a negative input.

Watch for: A larger base always gives a larger output for every x.

Extension: Compare 2ˣ and 3ˣ for x<0; explain why their ordering reverses.

Exit: Is (1/2)ˣ increasing or decreasing, and can its value be zero?

Decreasing; every finite real input has positive output.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-05 · 5–10 min · TEACHER · ANSWERS

A logarithm asks for an exponent

What number does log₂8 ask for, and why is log₂0 impossible?

Predict: Complete 2^?=8 and 2^?=1/4 before selecting logarithm inputs.

Compare	Computed observation
Positive logarithm Logarithm base a = 2; Argument b = 8	2^3 = 8, so log_2(8)=3.
Negative logarithm Logarithm base a = 2; Argument b = 0.25	2^-2 = 0.25, so log_2(0.25)=-2.
Zero argument Logarithm base a = 2; Argument b = 0	Require a>0, a≠1 and positive argument b.
Forbidden base 1 Logarithm base a = 1; Argument b = 8	Require a>0, a≠1 and positive argument b.

Conclusion: logₐb is the exponent y satisfying aʸ=b. Thus log₂8=3 and log₂(1/4)=−2. A positive exponential never equals zero or a negative number.

Conditions: A real logarithm requires a>0, a≠1 and argument b>0. A negative logarithm value is allowed.

Justification: The exponential aʸ is positive and strictly monotone for a>0,a≠1; it therefore has exactly one exponent for each positive output.

Ask: Always read logₐb aloud as “a to what exponent equals b?”.

Watch for: The logarithm value must be positive; logₐ0=0.

Extension: Explain why base 1 gives either no exponent or infinitely many, not a logarithm function.

Exit: Evaluate log₃(1/9).

−2 because 3⁻²=1/9.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-06 · 5–10 min · TEACHER · ANSWERS

Swap coordinates for inverse graphs

If (2,4) lies on y=2ˣ, which point lies on y=log₂x?

Predict: Reflect the point across y=x before plotting its inverse partner.

Compare	Computed observation
Point (2,4) Base a = 2; Exponential input x = 2	(2,4) reflects to (4,2).
Point with negative exponent Base a = 2; Exponential input x = -2	(-2,0.25) reflects to (0.25,-2).
Decreasing inverse pair Base a = 0.5; Exponential input x = -1	(-1,2) reflects to (2,-1).
Base 1 has no inverse Base a = 1; Exponential input x = 1	An invertible exponential needs a>0 and a≠1.

Conclusion: The relation y=aˣ is equivalent to x=logₐy. Swapping coordinates reflects the graph in y=x. Domain and range exchange: real inputs/positive outputs become positive inputs/real outputs.

Conditions: Require a>0,a≠1. The inverse graph has no point with input zero, even when its output is negative.

Justification: Applying logₐ to aˣ gives x and exponentiating logₐu gives u for u>0. These identities establish inversion beyond the displayed points.

Ask: Have students label each coordinate as input or output before swapping it.

Watch for: An inverse graph is obtained by changing y to −y.

Extension: Repeat for a=1/2 and input −1. Explain why both inverse functions decrease.

Exit: Reflect (−2,1/4) from y=2ˣ into its inverse graph.

(1/4,−2), on y=log₂x.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-07 · 5–10 min · TEACHER · ANSWERS

Products become sums of logarithms

Is logₐ(u+v)=logₐu+logₐv?

Predict: For base 2 and u=v=2, predict the logarithms of the product and of the sum.

Compare	Computed observation
Accidental equality at 2 and 2 Base a = 2; First argument u = 2; Second argument v = 2	Product rule: 2=2. Sum claim: 2 happens to equal 2; one accidental equality is not a general law.
Counterexample at 2 and 4 Base a = 2; First argument u = 2; Second argument v = 4	Product rule: 3=3. Sum claim: 2.5849625 differs from 3; one accidental equality is not a general law.
Positive quotient Base a = 2; First argument u = 8; Second argument v = 2	Product rule: 4=4. Sum claim: 3.3219281 differs from 4; one accidental equality is not a general law.
Positive product, forbidden factors Base a = 2; First argument u = -2; Second argument v = -4	Each argument u,v must be positive, even when their product is positive.

Conclusion: For positive u,v, logₐ(uv)=logₐu+logₐv. There is no corresponding sum rule. With u=2,v=4, the right side is 3 while log₂6 differs.

Conditions: Require a>0,a≠1 and u,v>0 separately. A positive product of two negative numbers does not make their separate real logarithms exist.

Justification: Write u=aʳ,v=aˢ. Then uv=a^(r+s), u/v=a^(r−s), and uᵏ=a^(kr). For valid base c, logₐu=log_cu/log_ca.

Ask: First use u=v=2 to expose an accidental agreement; then change v to 4 to demand a true counterexample.

Watch for: A rule for the logarithm of a product also works for a sum.

Extension: Use the product rule to derive quotient, power and change-of-base formulas, stating all domain conditions.

Exit: Evaluate log₂8−log₂2 and explain the single-log form.

2, equal to log₂(8/2)=log₂4.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-08 · 5–10 min · TEACHER · ANSWERS

The threshold between two integer rounds

When does 100rⁿ first reach a target, and why might rounding a logarithm fail?

Predict: For growth r=1.2 and target 150, predict the first whole round that reaches the target.

Compare	Computed observation
Growth to 150 Multiplier r = 1.2; Positive target = 150	First round: 3. Previous value 144; at round 3: 172.8.
Decay to 50 Multiplier r = 0.8; Positive target = 50	First round: 4. Previous value 51.2; at round 4: 40.96.
Target at the initial value Multiplier r = 1.2; Positive target = 100	First round: 0. at round 0: 100.
Constant cannot reach 150 Multiplier r = 1; Positive target = 150	The constant model cannot reach this target in the stated direction.
Nonpositive target rejected Multiplier r = 0.8; Positive target = 0	The multiplier and target must both be positive.

Conclusion: For growth r>1 and target above 100, the real crossing is n*=log(target/100)/log r, but the first integer round is the least integer ≥n*. For decay 0<r<1, use the first round at or below the target.

Conditions: n is a nonnegative integer and the target is positive. The question is at-or-above for growth and at-or-below for decay. At r=1 only a target already met at n=0 is reachable.

Justification: Monotonicity converts the threshold to an inequality in n; evaluate the candidate round and its predecessor to certify first attainment. A logarithmic approximation alone is insufficient.

Ask: Require both a failed preceding round and a successful first round in the written explanation.

Watch for: Round the real logarithmic answer to the nearest integer.

Extension: For r=0.8 and target 50, check the two integer rounds on either side of the real crossing.

Exit: 100×1.2ⁿ reaches at least 150 first at which integer n?

3: round 2 gives 144 and round 3 gives 172.8.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

06 Exponential and logarithmic functions · g10-06-09 · 5–10 min · TEACHER · ANSWERS

A decreasing base reverses the inequality

Does logₐx<logₐb always imply x<b?

Predict: Solve log₂x<log₂4 and log_(1/2)x<log_(1/2)4 before testing a value.

Compare	Computed observation
Increasing logarithm Function family = Logarithmic comparison; Base a = 2; Test input x = 2; Comparison input b = 4	Solution: 0 < x < 4. Current input satisfies the strict inequality.
Decreasing logarithm, same inputs Function family = Logarithmic comparison; Base a = 0.5; Test input x = 2; Comparison input b = 4	Solution: x > 4. Current input does not satisfy the strict inequality.
Equality boundary Function family = Logarithmic comparison; Base a = 0.5; Test input x = 4; Comparison input b = 4	Solution: x > 4. Current input does not satisfy the strict inequality.
Forbidden logarithm input zero Function family = Logarithmic comparison; Base a = 2; Test input x = 0; Comparison input b = 4	Require a>0,a≠1; logarithms additionally need both arguments positive.
Increasing exponential at negative input Function family = Exponential comparison; Base a = 2; Test input x = -1; Comparison input b = 0	Solution: x < 0. Current input satisfies the strict inequality.
Decreasing exponential, same inputs Function family = Exponential comparison; Base a = 0.5; Test input x = -1; Comparison input b = 0	Solution: x > 0. Current input does not satisfy the strict inequality.

Conclusion: For a>1, logₐx<logₐb gives 0<x<b and aˣ<aᵇ gives x<b. For 0<a<1, both functions decrease and both comparisons instead give x>b. Equality x=b is excluded.

Conditions: Require a>0,a≠1. Logarithmic arguments x,b must be positive; exponential inputs x,b may be any real numbers. Base 1 is excluded here.

Justification: Strict monotonicity gives equivalence with x<b for an increasing log and x>b for a decreasing log. The positive-argument restriction must be intersected with the result.

Ask: Keep the same test input while switching only the base from 2 to 1/2.

Watch for: Taking logarithms always preserves the inequality direction.

Extension: State the corresponding rule for aˣ<aᵇ and compare the domain restrictions.

Exit: Solve log_(1/2)x<log_(1/2)4.

x>4.

Before: Real-number operations and reading coordinates; write a prediction on paper. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.