What is the value of log₃ 81?
- 27
- 4 Correct answer.
- 9
- 3
Explanation
The logarithm log₃ 81 asks for the exponent x such that 3^x = 81. Since 3⁴ = 81, the value is 4.
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The logarithm log₃ 81 asks for the exponent x such that 3^x = 81. Since 3⁴ = 81, the value is 4.
For a quadratic with leading coefficient 1, the roots must add to the coefficient of x with opposite sign (5) and multiply to the constant term (6). The pair 2 and 3 satisfies both conditions, since (x − 2)(x − 3) expands to x² − 5x + 6.
The Pythagorean identity states that the squares of the sine and cosine of the same angle always sum to 1. This follows from the definitions of sine and cosine on a unit circle of radius 1.
The area of a circle is πr². Substituting r = 5 gives π(5)² = 25π square centimeters.
Write i³ as i² · i and substitute i² = −1. This gives (−1)(i) = −i.
The first term is 3 and the common difference is 4. Using aₙ = a₁ + (n − 1)d with n = 10 gives 3 + 9(4) = 39.
The power rule states that the derivative of axⁿ is anxⁿ⁻¹. Here a = 3 and n = 2, so the derivative is 3 · 2 · x¹ = 6x.
A polynomial is continuous everywhere, so the limit equals the function value at x = 3. Substituting gives 4(3) − 5 = 7.
By the binomial theorem, the term containing x³ is C(5,3)x³(2)². This gives 10 · x³ · 4 = 40x³, so the coefficient is 40.
The word has 5 letters with L repeated twice and E repeated twice. The number of distinct arrangements is 5!/(2!·2!) = 120/4 = 30.
An antiderivative of 2x + 1 is x² + x. Evaluating from 0 to 1 gives (1 + 1) − (0 + 0) = 2.
For a 2×2 matrix with entries a, b in the first row and c, d in the second, the determinant is ad − bc. Here that is (3)(5) − (4)(2) = 15 − 8 = 7.
The prime outcomes on a die are 2, 3, and 5, giving three favorable outcomes out of six equally likely ones. The probability is 3/6 = 1/2.
Apply the identity sin²θ + cos²θ = 1: cos²θ = 1 − 9/25 = 16/25. In the first quadrant cosine is positive, so cos θ = 4/5.
Compound interest follows A = P(1 + r)ᵗ. With r = 0.10 and t = 2, A = P(1.10)² = 1.21P.
Use the point-slope form y − y₁ = m(x − x₁) with m = 2 and (x₁, y₁) = (2, 3). This gives y − 3 = 2(x − 2), which simplifies to y = 2x − 1.
The dot product is the sum of the products of corresponding components: (2)(1) + (−1)(4) + (3)(2) = 2 − 4 + 6 = 4.
Here the first term is 4 and the common ratio is 1/2, so the sum is a/(1 − r) = 4/(1 − 1/2) = 8.
By Vieta's formulas, the sum of the roots of ax² + bx + c = 0 is −b/a. Here that is −(−8)/2 = 4.
First square: (1 + i)² = 1 + 2i + i² = 2i. Squaring again gives (2i)² = 4i² = −4.
Separating variables gives dy/y = 2 dx, so ln y = 2x + C and y = Ae^(2x). Applying y(0) = 3 gives A = 3, hence y = 3e^(2x).
Multiplication by e^(at) shifts the transform of the base function by a, replacing s with s − a. Since the transform of sin 3t is 3/(s² + 9), the result is 3/((s − 2)² + 9).
The total number of 3-person committees is C(9,3) = 84, and the number made up entirely of women is C(4,3) = 4. The probability is therefore 4/84 = 1/21.
Setting f′(x) = 12 − 3x² = 0 gives the critical point x = 2 inside the interval. Comparing f(0) = 0, f(2) = 16, and f(3) = 9, the largest value is 16.
The z-score is (86 − 70)/8 = 2. The area to the right of z = 2 under the standard normal curve is 0.0228.
A logarithm answers the question of what exponent on the base gives the argument. Since 2⁵ = 32, log₂ 32 = 5.
The modulus of a + bi is √(a² + b²). Here √(3² + (−4)²) = √25 = 5.
⁷C₃ = 7!/(3!·4!) = (7·6·5)/(3·2·1) = 35. It counts the 3-element subsets of a 7-element set.
The chain rule requires multiplying by the derivative of the exponent. Since d/dx(3x) = 3, the derivative is 3e^(3x).
Perpendicular slopes are negative reciprocals, so their product is −1. The negative reciprocal of 2/3 is −3/2.
sin 30° = 1/2 and cos 60° = 1/2 because the two angles are complementary. Their sum is therefore 1.
An arithmetic series with first term 1 and last term 20 sums to n(n + 1)/2. Substituting n = 20 gives 20·21/2 = 210.
The common ratio is 6/2 = 3, and the nth term is a₁r^(n−1). Thus a₅ = 2·3⁴ = 2·81 = 162.
Rewrite the expression as 5·sin(5x)/(5x) and use the standard limit sin u/u → 1 as u → 0. The result is 5·1 = 5.
Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫e^u du = (1/2)e^(x²) + C.
The cross product is (1·0 − 0·3, 0·1 − 2·0, 2·3 − 1·1) = (0, 0, 5). Its magnitude is 5, equal to the area of the parallelogram formed by A and B.
Conditional probability is defined as P(A | B) = P(A ∩ B)/P(B). Dividing 0.2 by 0.4 gives 0.5.
The larger denominator 16 lies under y², so the major axis is vertical with a² = 16 and b² = 9. Then c² = a² − b² = 7, giving foci (0, ±√7).
Adding the two equations eliminates y and gives 3x = 9, so x = 3. Substituting back gives 6 + y = 7, so y = 1.
The mean is 6, and the squared deviations are 16, 4, 0, 4, 16. Their average is 40/5 = 8, which is the variance.
A p-series Σ 1/n^p converges exactly when p > 1. Only the exponent 2 exceeds 1, so that is the convergent series.
The general term is ⁶C_k (2x)^k. For k = 2 this gives ⁶C₂·4 = 15·4 = 60.
The dot product is 1 and both vectors have magnitude √2 and 1 respectively. So cos θ = 1/√2, giving θ = 45°.
The antiderivative is arctan x. Evaluating from 0 to 1 gives arctan 1 − arctan 0 = π/4.
Multiplying through gives 1 = A(x + 1) + B(x − 1). Setting x = 1 yields 1 = 2A, so A = 1/2.
Integrating gives [−1/x] from 1 to b, which equals 1 − 1/b. Taking the limit as b → ∞ leaves 1, so the integral converges.
Completing the square gives (s + 2)² + 1, which matches the shifted form 1/((s − a)² + ω²). With a = −2 and ω = 1, the inverse is e^(−2t) sin t.
The characteristic equation r² − 3r + 2 = 0 has roots 1 and 2, so y = Ae^t + Be^(2t). The conditions give A + B = 1 and A + 2B = 0, hence A = 2 and B = −1.
The curve r = 2cos θ traces a circle of radius 1 centered at (1, 0) in the plane. Its enclosed area is therefore π·1² = π, consistent with the formula (1/2)∫r² dθ over one full trace.
Solve det(A − λI) = (2 − λ)² − 1 = 0, which factors as (1 − λ)(3 − λ) = 0. The eigenvalues are therefore 1 and 3.