7 If logs(x) = 2, find x. 8 Solve: 3x+1 = 81 9 A bacterial culture doubles every hour. If there are initially 100 bacteria, how many will there be after 5 hours? 10 The pH of a solution is given by pH = -log10[H+]. If [H+] = 10-7, what is the pH?

Problem 7: If _5(x) = 2, find x. (Assuming the base of the logarithm is 5, as this leads to one of the options.) Step 1: Convert the logarithmic equation to an exponential equation. _5(x) = 2 x = 5^2 Step 2: Calculate the value of x. x = 25 The correct option is D. D. x = 25 Problem 8: Solve: 3^x+1 = 81. Step 1: Express 81 as a power of 3. 81 = 3^4 Step 2: Equate the exponents since the bases are the same. 3^x+1 = 3^4 x+1 = 4 Step 3: Solve for x. x = 4 - 1 x = 3 The correct option is B. B. x = 3 Problem 9: A bacterial culture doubles every hour. If there are initially 100 bacteria, how many will there be after 5 hours? Step 1: Identify the initial number of bacteria and the growth factor. Initial bacteria P_0 = 100. The culture doubles every hour, so the growth factor is 2. The number of hours is t = 5. Step 2: Use the exponential growth formula P(t) = P_0 · 2^t. P(5) = 100 · 2^5 Step 3: Calculate the value. P(5) = 100 · 32 P(5) = 3200 The correct option is C. C. 3200 Problem 10: The pH of a solution is given by pH = -_10[H^+]. If [H^+] = 10^-7, what is the pH? Step 1: Substitute the given concentration of hydrogen ions into the pH formula. pH = -_10(10^-7) Step 2: Use the logarithm property _b(b^x) = x. pH = -(-7) Step 3: Simplify the expression. pH = 7 The correct option is C. C. 7 Problem 11: Simplify: (2^-3 × 2^5) ÷ 2^2. Step 1: Apply the product rule for exponents (a^m × a^n = a^m+n) to the terms inside the parenthesis. 2^-3 × 2^5 = 2^-3+5 = 2^2 Step 2: Substitute this result back into the expression. 2^2 ÷ 2^2 Step 3: Apply the quotient rule for exponents (a^m ÷ a^n = a^m-n). 2^2 ÷ 2^2 = 2^2-2 = 2^0 Step 4: Simplify the expression. 2^0 = 1 The correct option is A. A. 1 Problem 12: Find the remainder when 47 is divided by 5. Step 1: Perform the division of 47 by 5. 47 = 5 × q + r Step 2: Determine the quotient (q) and the remainder (r). 47 = 5 × 9 + 2 The remainder is 2. The correct option is B. B. 2 Problem 13: Solve: 3x 2 5. Step 1: Find the multiplicative inverse of 3 modulo 5. We need a number k such that 3k 1 5. 3 × 2 = 6 1 5 So, the inverse of 3 modulo 5 is 2. Step 2: Multiply both sides of the congruence by the inverse (2). 2 × (3x) 2 × 2 5 6x 4 5 Step 3: Simplify the congruence. Since 6 1 5. 1x 4 5 x 4 5 The correct option is D. D. x 4 ±od5 Problem 14: What is 23 7? Step 1: Divide 23 by 7. 23 = 7 × q + r Step 2: Determine the quotient (q) and the remainder (r). 23 = 7 × 3 + 2 The remainder is 2. The correct option is B. B. 2 Problem 15: Solve: 2x + y = 7 and x - y = 2. Step 1: Use the elimination method by adding the two equations. (2x + y) + (x - y) = 7 + 2 3x = 9 Step 2: Solve for x. x = (9)/(3) x = 3 Step 3: Substitute the value of x into the second equation (x - y = 2) to find y. 3 - y = 2 y = 3 - 2 y = 1 The solution is x=3, y=1. The correct option is A. A. x=3, y=1 Problem 16: Solve: 3x + 2y = 12 and x + y = 5. Step 1: From the second equation, express y in terms of x. y = 5 - x Step 2: Substitute this expression for y into the first equation. 3x + 2(5 - x) = 12 Step 3: Solve for x. 3x + 10 - 2x = 12 x + 10 = 12 x = 12 - 10 x = 2 Step 4: Substitute the value of x back into the expression for y. y = 5 - 2 y = 3 The solution is x=2, y=3. The correct option is A. A. x=2, y=3 Send me the next one 📸