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Among the following functions, for which one is f(a – b) = f(a) * f(–b) for all integer values of a and b?
A. f(t) = t/10
B. f(t) = t + 10
C. f(t) = 10t
D. f(t) = 10^t
E. f(t) = t^10
A. f(t) = t/10
B. f(t) = t + 10
C. f(t) = 10t
D. f(t) = 10^t
E. f(t) = t^10
09 Apr 2026, 06:02
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Among the following functions, for which one is f(a – b) = f(a) * f(–b) for all integer values of a and b?
Algebraic Approach
A. f(t) = t/10
(a - b)/10 ≠ -ab/100
B. f(t) = t + 10
(a - b) + 10 ≠ (a + 10)(-b + 10)
C. f(t) = 10t
10(a - b) ≠ (10a)(-10b)
D. f(t) = 10^t
10^(a - b) = (10^a)(10^-b)
E. f(t) = t^10
(a - b)^10 ≠ (a^10)(-b^10)
Smart Numbers Approach (very fast)
Let a and b be 1.
A. f(t) = t/10
0/10 ≠ -1/100
B. f(t) = t + 10
0 + 10 ≠ (1 + 10)(-1 + 10)
C. f(t) = 10t
10 * 0 ≠ 10 * -10
D. f(t) = 10^t
10^0 = (10^1)(10^-1)
E. f(t) = t^10
0^10 ≠ (1^10)(-1^10)
Also, we can save time by noticing that, since the left side of the equation involves subtraction whereas the right side involves multiplication, the correct answer must be in a form such that subtraction and multiplication are related. Having seen that, we can try (D) first because subtraction of exponents is related to multiplying positive and negative exponents.
Correct answer: D
Algebraic Approach
A. f(t) = t/10
(a - b)/10 ≠ -ab/100
B. f(t) = t + 10
(a - b) + 10 ≠ (a + 10)(-b + 10)
C. f(t) = 10t
10(a - b) ≠ (10a)(-10b)
D. f(t) = 10^t
10^(a - b) = (10^a)(10^-b)
E. f(t) = t^10
(a - b)^10 ≠ (a^10)(-b^10)
Smart Numbers Approach (very fast)
Let a and b be 1.
A. f(t) = t/10
0/10 ≠ -1/100
B. f(t) = t + 10
0 + 10 ≠ (1 + 10)(-1 + 10)
C. f(t) = 10t
10 * 0 ≠ 10 * -10
D. f(t) = 10^t
10^0 = (10^1)(10^-1)
E. f(t) = t^10
0^10 ≠ (1^10)(-1^10)
Also, we can save time by noticing that, since the left side of the equation involves subtraction whereas the right side involves multiplication, the correct answer must be in a form such that subtraction and multiplication are related. Having seen that, we can try (D) first because subtraction of exponents is related to multiplying positive and negative exponents.
Correct answer: D
09 Apr 2026, 06:33
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The key is spotting what mathematical structure lets subtraction in the input "match" multiplication in the output. Most people test all five options, but there's a faster path.
Notice the left side has a minus sign in the argument (a - b), while the right side has separate function values multiplied together. For that to work, the function needs to convert subtraction into division — and the only function type that does this is an exponential, via the exponent rule x^(a-b) = x^a * x^(-b).
1. Set up the condition: f(a - b) = f(a) * f(-b) for all integers a and b.
2. Test choice D: f(t) = 10^t.
Left side: f(a - b) = 10^(a - b).
Right side: f(a) * f(-b) = 10^a * 10^(-b) = 10^(a - b).
Both sides are equal. This works for all integers.
3. Why the others fail:
A. f(t) = t/10. Left: (a-b)/10. Right: (a/10) * (-b/10) = -ab/100. Not the same.
B. f(t) = t + 10. Left: a - b + 10. Right: (a + 10)(-b + 10). Not the same.
C. f(t) = 10t. Left: 10(a - b). Right: (10a)(-10b) = -100ab. Not the same.
E. f(t) = t^10. Left: (a - b)^10. Right: a^10 * (-b)^10 = a^10 * b^10. These are only equal for specific values, not all integers.
The small trap in D is thinking that f(-b) = 10^(-b) is somehow problematic. It's not. Negative exponents are fine — 10^(-b) is just 1/10^b, and the exponent rule still applies cleanly.
Shortcut if you're pressed for time: as soon as you recognize that the left side has (a - b) inside a function and the right side is a product, think "exponents turn subtraction into division" and go straight to D. That's the only function family where this holds.
Answer: D
Notice the left side has a minus sign in the argument (a - b), while the right side has separate function values multiplied together. For that to work, the function needs to convert subtraction into division — and the only function type that does this is an exponential, via the exponent rule x^(a-b) = x^a * x^(-b).
1. Set up the condition: f(a - b) = f(a) * f(-b) for all integers a and b.
2. Test choice D: f(t) = 10^t.
Left side: f(a - b) = 10^(a - b).
Right side: f(a) * f(-b) = 10^a * 10^(-b) = 10^(a - b).
Both sides are equal. This works for all integers.
3. Why the others fail:
A. f(t) = t/10. Left: (a-b)/10. Right: (a/10) * (-b/10) = -ab/100. Not the same.
B. f(t) = t + 10. Left: a - b + 10. Right: (a + 10)(-b + 10). Not the same.
C. f(t) = 10t. Left: 10(a - b). Right: (10a)(-10b) = -100ab. Not the same.
E. f(t) = t^10. Left: (a - b)^10. Right: a^10 * (-b)^10 = a^10 * b^10. These are only equal for specific values, not all integers.
The small trap in D is thinking that f(-b) = 10^(-b) is somehow problematic. It's not. Negative exponents are fine — 10^(-b) is just 1/10^b, and the exponent rule still applies cleanly.
Shortcut if you're pressed for time: as soon as you recognize that the left side has (a - b) inside a function and the right side is a product, think "exponents turn subtraction into division" and go straight to D. That's the only function family where this holds.
Answer: D
