If log₂(x) = 5, then x = ? — 32.log₂(x) = 5 → x = 2⁵ = 32.
To solve 5^x = 125, you recognize that 125 = 5^? — 3.5³ = 125. So 5^x = 5³ → x = 3.
Solve: 3^x = 81 — x = 4.3⁴ = 81, so x = 4.
Solve: 2^(x+1) = 32 — x = 4.32 = 2⁵, so x + 1 = 5 → x = 4.
To solve 5^x = 7 (where we cannot express 7 as a power of 5), the first step is: — Take log of both sides.Take log (or ln) of both sides: x·log(5) = log(7) → x = log(7)/log(5).
Solve: e^x = 20 — x = ln(20).Take ln of both sides: x = ln(20) ≈ 2.996.
Solve: log₃(x) = 4 — x = 81.log₃(x) = 4 → x = 3⁴ = 81.
Solve: 6^x = 1 — x = 0.Any nonzero base raised to power 0 equals 1. So 6^x = 1 → x = 0.
The value of e is approximately: — 2.71828.e ≈ 2.71828… It is the base of the natural logarithm and arises in continuous growth.
The natural logarithm ln(x) is log base: — e.ln(x) = log_e(x). The natural logarithm uses base e.
In the model A(t) = A₀eʳᵗ, positive r means: — Growth.When r > 0, the exponent rt grows with time, so A(t) increases — this is growth.
In the model A(t) = A₀eʳᵗ, negative r means: — Decay.When r < 0, rt becomes more negative over time, so e^(rt) decreases — this is decay.
If a substance has half-life 5 years, after 10 years the remaining amount is: — 1/4 of original.Two half-lives pass: (1/2)² = 1/4 of the original remains.
ln(e) = ? — 1.ln(e) = log_e(e) = 1. Just as log₁₀(10) = 1.
e^0 = ? — 1.Any number to the power 0 equals 1. e^0 = 1.
Which graph best describes y = e^(−x)? — Starts high, decreases toward y = 0.e^(−x) is a decreasing exponential. As x→∞, e^(−x)→0 (asymptote at y=0). At x=0, y=1.
For A(t) = A₀e^(rt), what is the y-intercept? — A₀.At t = 0: A(0) = A₀e^0 = A₀. The initial value A₀ is the y-intercept.