SOLVING LOGARITHMIC AND INDICES PROBLEM Solving equation in

SOLVING LOGARITHMIC AND INDICES PROBLEM

Solving equation in the form of ax = ay If ax = ay then x = y Example: 32 x = 27 = 33 By comparing index: 2 x = 3

Examples 8 x+1 = 4 x+3 (23)x+1 = (22)x+3 23 x+3 = 22 x+6 By comparing index: 3 x + 3 = 2 x + 6 3 x – 2 x = 6 – 3 x=3

Examples 9 x. 3 x 1 = 243 32 x. 3 x 1 = 35 32 x + (x 1) = 35 33 x 1 = 35 By comparing index, 3 x – 1 = 5 3 x = 6 x=2

A very different example. SOLVE 2 X + 2 X+3 = 32 2 X + 2 X+3 = 25 x+x+3=5 2 x = 2 x=1 WARNING! The solution above is WRONG!!! See the right way ---->

SOLVE 2 X + 2 X+3 = 32 2 x + 2 x 23 = 32 Factorize 2 x 2 X(1 + 23) = 32 2 X (9)= 32 2 X = 32/9 X lg 2 = lg 32 – lg 9 X (0. 3010)= 1. 5051 -0. 9542 X=1. 8302

Solving equation in the form of ax = b, where a ≠-1, 0 , 1 INDEX EQUATION WITH DIFFERENT BASE If we cannot express both sides of the equation with the same base, we solve the equation by taking logarithms on both sides. Example 5 x= 6 Taking logarithms on both sides. log 10 5 x = log 10 6 x log 10 5 = log 10 6 x (0. 6990) = 0. 7782 x = 0. 7782 0. 6690 x = 1. 113

Example: Solve 5 x – 3 x+1 = 0 Solution: 5 x – 3 x+1 = 0 5 x = 3 x+1 Taking logarithms on both sides, lg 5 x = lg 3 x+1 x lg 5 = (x + 1) lg 3 x lg 5 = x lg 3 + lg 3 x lg 5 – x lg 3= lg 3 x(lg 5 – lg 3)= lg 3 x(0. 6990 – 0. 4771) = 0. 4771 x = 2. 150

Solving Logarithmic Equation Solve log 5 (5 x – 4) = 2 log 5 3 + log 5 4 First, simplify the right hand side. log 5 (5 x – 4) = 2 log 5 3 + log 5 4 log 5 (5 x – 4 ) = log 5 (36) Comparing number in both sides. log 5 (5 x – 4) = log 5 (36) 5 x = 40 x=8

Solve the equation log 5 x = 4 logx 5 Solution: (Change base from x to 5) log 5 x = 4 (log 5 x)2 = 4 log 5 x = 2 or -2 x = 52 or 5 2