answer:First, convert the logarithmic equation to its exponential equivalent: log_{64} (3x - 2) = -frac{1}{3} This means: 3x - 2 = 64^{-frac{1}{3}} Since 64 = 2^6, you can write: 3x - 2 = (2^6)^{-frac{1}{3}} = 2^{-2} = frac{1}{4} Now solve for x: 3x - 2 = frac{1}{4} 3x = frac{1}{4} + 2 3x = frac{1}{4} + frac{8}{4} = frac{9}{4} x = frac{9}{4} times frac{1}{3} = frac{9}{12} = frac{3}{4} Conclusion: Thus, the solution to the problem is x = boxed{frac{3}{4}}.

answer:First, let's calculate the total cost for each type of light bulb Valerie needs to buy: For Lamp A: 2 small light bulbs at 8.75 each = 2 * 8.75 = 17.50 For Lamp B: 3 medium light bulbs at 11.25 each = 3 * 11.25 = 33.75 For Lamp C: 1 large light bulb at 15.50 each = 1 * 15.50 = 15.50 For Lamp D: 4 extra-small light bulbs at 6.10 each = 4 * 6.10 = 24.40 For Lamp E: 2 large light bulbs at 15.50 each = 2 * 15.50 = 31.00 For Lamp F: 1 small light bulb at 8.75 each and 1 medium light bulb at 11.25 each = 8.75 + 11.25 = 20.00 Now, let's add up the total cost for all the light bulbs: Total cost = 17.50 (Lamp A) + 33.75 (Lamp B) + 15.50 (Lamp C) + 24.40 (Lamp D) + 31.00 (Lamp E) + 20.00 (Lamp F) Total cost = 142.15 Since Valerie has a budget of 120, she will not have enough money to buy all the light bulbs she needs. She will be short by 142.15 - 120 = boxed{22.15} .

answer:1. **Define the polynomial and sequence:** Let ( P(x) = x^2 + 1 ). Define the sequence ( a_i = P^i(0) ) for all ( i geq 1 ). This means: [ a_1 = P(0) = 0^2 + 1 = 1 ] [ a_2 = P(a_1) = P(1) = 1^2 + 1 = 2 ] [ a_3 = P(a_2) = P(2) = 2^2 + 1 = 5 ] and so on. 2. **Inductive growth of the sequence:** We observe that ( a_{n+1} > a_n ) for all ( n geq 1 ). By induction, we can show that ( a_{n+1} > a_n^2 > a_1 a_2 ldots a_n ) for all ( n geq 2 ). 3. **Existence of a prime divisor:** Fix a positive integer ( n ). We need to show that there exists a prime divisor ( p ) of ( a_n ) that does not divide any of the previous ( a_i ) for ( i < n ). This implies that there is a cycle of length ( n ) for that prime ( p ). 4. **Prime divisor properties:** There exists a prime ( p ) such that ( nu_p(a_n) > nu_p(a_1 a_2 ldots a_{n-1}) ). Suppose otherwise, and let ( p mid P^k(0) ) where ( k ) is minimal. Then ( k mid n ). 5. **Valuation of the sequence:** We will show that ( nu_p(a_{mk}) = nu_p(a_k) ) for all positive integers ( m ), which will contradict the choice of ( p ). Observe that the polynomial ( P^{mk}(x) ) does not have an ( x^1 ) term, so: [ nu_p(P^{(m+1)k}(0)) = nu_p(P^{mk}(a_k)) = min(nu_p(P^{mk}(0)), 2nu_p(a_k)) ] For ( m = 1 ), it is ( nu_p(a_k) ), and for ( m geq 2 ), by induction, it remains the same. Thus, we have a contradiction. 6. **Conclusion:** Since we can always find such a prime ( p ) for any ( n ), it implies that ( f(p) ) can attain arbitrarily large values. (blacksquare)

answer:1. **Calculate the distance from -3 to -8**: [ |-8 - (-3)| = |-8 + 3| = |-5| = 5 text{ units} ] 2. **Calculate the distance from -8 to 2**: [ |2 - (-8)| = |2 + 8| = |10| = 10 text{ units} ] 3. **Calculate the distance from 2 to 7**: [ |7 - 2| = |5| = 5 text{ units} ] 4. **Add all distances for the total distance**: [ 5 text{ units} + 10 text{ units} + 5 text{ units} = 20 text{ units} ] The total distance the bug crawls is 20 text{ units}. The correct answer is boxed{textbf{(B)} 20}.