a) Solve the equation $$5\log_3(3x-1)=10.$$ (2 marks)b) Express $y$ in terms of $x$ in a way that not include logarithms $$\log_x(15y-5)-\log_x(20)=7.$$ (3 marks)c) Solve the equation giving your solutions to 2 decimal places$$2(7^x)-3(7^{-x})+5=0.$$ (5 marks)

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This question contains three subparts, each requiring a distinct solving approach.
Part a) Start by isolating the logarithm. Given 5 log_3(3x-1) = 10, divide both sides by 5 to get log_3(3x-1) = 2. By the definition of logarithms, if log_b(A) = C then A = b^C. Here, 3x - 1 = 3^2 = 9. Solve for x: 3x = 10, so x = 10/3 ≈ 3.333...

Part b) Use log subtraction rules to simplify. The expression log_x(15y-5) - log_x(20) equals log_x((15y-5)/20) by the property log_b(M) - log_b(N) = log_b(M/N). Set this equal to 7, so log_x((15y-5)/20) = 7. Convert the logarithmic equation to an exponential one: x^7 = (15y-5)/20. Multiply both sides by 20: 20x^7 = 15y - 5. Then solve for y: 15y = 20x^7 + 5, so y = (20x^7 + 5)/15 = (4x^7)/3 + 1/3.

Part c) Let t = 7^x (note t > 0). The equation becomes 2t - 3/t + 5 = 0. Multiply through by t to clear the fraction: 2t^2 + 5t - 3 = 0, a standard quadratic in t. Compute the discriminant: Δ = 5^2 - 4(2)(-3) = 25 + 24 = 49, so √Δ = 7. The roots are t = [-5 ± 7]/(2·2) = [-5 ± 7]/4. This yields t = 2/4 = 1/2 or t = -12/4 = -3. Since t = 7^x > 0, discard t = -3 and accept t = 1/2. Now solve 7^x = 1/2 by taking logarithms: x = log(1/2) / log(7). Numerically, x ≈ -0.3562, which rounds to -0.36 to two decimal places.

FC311 KICL 8.5 Exponential and Logarithms Class Test AY 24-25

a) Solve the equation \[5\log_3(3x-1)=10.\] (2 marks)b) Express \(y\) in terms of \(x\) in a way that not include logarithms \[\log_x(15y-5)-\log_x(20)=7.\] (3 marks)c) Solve the equation giving your solutions to 2 decimal places\[2(7^x)-3(7^{-x})+5=0.\] (5 marks)

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