If logb2 = x and logb3 = y, evaluate the following in terms of x and y: You need these three rules: 1. logb(ac) = logba + logbc 2. logb = logba - logbc 3. logb = c·logba (a) logb216= Break 216 down to primes: 216 = = = = = = logb216 = logb = logb + logb (by rule 1.) = 3·logb2 + 3·logb3 (by rule 3.) = 3·x + 3·y (by substitution.) = 3(x + y) ============================================= (b) logb162= Break 162 down to primes: 162 = = = = logb162 = logb = logb2 + logb (by rule 1.) = logb2 + 4·logb3 (by rule 3.) = x + 4·y (by substitution.) = x + 4y ======================================= (c) logb Break the 9 in the denominator down into primes 9 = = logb = logb = logb2 - logb = logb2 - 2·logb3 = x - 2y ============================================ (d) logb9 ——————— logb8 Break the 9 down as and the 8 down as logb9 logb ——————— = ————————— logb8 logb 2·logb3 ————————— (by rule 3.) 3·logb2 (substitution) Edwin
