Question 1

What is an exponential equation?

Accepted Answer

An exponential equation is an equation in which the variable appears in the exponent rather than the base. Examples: 2ˣ = 32, 5^(2x−1) = 125, eˣ = 7. These differ from polynomial equations where the variable is in the base (x² = 9). Because exponential functions grow (or decay) very rapidly, they model many real-world phenomena including compound interest (A = P·eʳᵗ), population growth, carbon-14 dating, and the spread of disease.

Question 2

How do you solve aˣ = b using logarithms?

Accepted Answer

Take the logarithm of both sides (any base works; base 10 or base e are most convenient). Left side: log(aˣ) = x·log(a) by the power rule. Right side: log(b). So x·log(a) = log(b), giving x = log(b)/log(a). Example: 3ˣ = 20 → x = log(20)/log(3) = 1.301/0.477 ≈ 2.727. You can verify: 3^2.727 ≈ 20 ✓. Using natural log gives the same result: x = ln(20)/ln(3) ≈ 2.996/1.099 ≈ 2.727.

Question 3

How do you solve eˣ = b?

Accepted Answer

When the base is e (Euler's number ≈ 2.718), take the natural logarithm (ln) of both sides. Since ln(eˣ) = x by definition, the equation simplifies immediately: eˣ = b → x = ln(b). Example: eˣ = 50 → x = ln(50) ≈ 3.912. For eˣ = 1 the answer is x = 0; for eˣ = e the answer is x = 1. If b ≤ 0 there is no real solution because eˣ is always positive.

Question 4

How do you solve 2ˣ = 16 step by step?

Accepted Answer

Method 1 — Common base: Recognize that 16 = 2⁴. So 2ˣ = 2⁴, which means x = 4. Method 2 — Logarithms: Take log of both sides: x·log(2) = log(16), so x = log(16)/log(2) = 1.2041/0.3010 = 4. Both methods confirm x = 4. Verification: 2⁴ = 16 ✓. Use Method 1 when the right side is a perfect power of the base; use Method 2 (logarithms) when it is not.

Question 5

How are exponential equations used in growth and decay problems?

Accepted Answer

Exponential growth is modeled by A = A₀·eᵏᵗ (k > 0) and exponential decay by A = A₀·e^(−kt) (k > 0). To find the time t at which A reaches a target value, isolate eᵏᵗ and take the natural log: t = ln(A/A₀)/k. For example, a bacteria population doubles every 3 hours starting at 500: 500·e^(k·3) = 1000, so k = ln(2)/3 ≈ 0.231. To reach 8000: 500·e^(0.231·t) = 8000, t = ln(16)/0.231 ≈ 12 hours.

Equation	Method	Solution	Example
aˣ = aⁿ	Equate exponents	x = n	2ˣ = 2⁴ → x = 4
aˣ = b	Take log both sides	x = log(b)/log(a)	3ˣ = 20 → x ≈ 2.727
eˣ = b	Take ln both sides	x = ln(b)	eˣ = 50 → x ≈ 3.912
10ˣ = b	Take log₁₀ both sides	x = log(b)	10ˣ = 500 → x ≈ 2.699
e^(kx) = b	ln both sides	x = ln(b)/k	e^(2x) = 10 → x ≈ 1.151
A₀·eᵏᵗ = A	Isolate eᵏᵗ, ln both sides	t = ln(A/A₀)/k	500·e^(0.1t)=2000 → t=13.86
a^(f(x)) = a^(g(x))	Equate f(x)=g(x)	Solve f(x)=g(x)	2^(x²) = 2^(3x−2) → x=1 or 2

Equation	log(b)	log(a)	x = log(b)/log(a)	Verification
2ˣ = 32	log(32) = 1.505	log(2) = 0.301	1.505/0.301 = 5	2⁵ = 32 ✓
3ˣ = 81	log(81) = 1.908	log(3) = 0.477	1.908/0.477 = 4	3⁴ = 81 ✓
5ˣ = 125	log(125) = 2.097	log(5) = 0.699	2.097/0.699 = 3	5³ = 125 ✓
2ˣ = 15	log(15) = 1.176	log(2) = 0.301	1.176/0.301 ≈ 3.907	2^3.907 ≈ 15 ✓
10ˣ = 1000	log(1000) = 3	log(10) = 1	3/1 = 3	10³ = 1000 ✓
eˣ = 1	ln(1) = 0	ln(e) = 1	0/1 = 0	e⁰ = 1 ✓

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