1. SOLVING LOGARITHMIC AND INDICES PROBLEM

2. Solving equation in the form of a x = a y Example: 3 2x = 27 = 3 3 By comparing index: 2x = 3 If a x = a y then x = y

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

4. 9 x. 3 x  1 = 243 3 2x. 3 x  1 = 3 5 3 2x + (x  1) = 3 5 3 3x  1 = 3 5 By comparing index, 3x – 1 = 5 3x = 6 x = 2

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

6. 2 x + 2 x  2 3 = 32 Factorize 2 x 2 X (1 + 2 3 ) = 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 SOLVE 2 X + 2 X+3 = 32

7. 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 Solving equation in the form of a x = b, where a ≠-1, 0, 1

8. 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

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

10. Solve the equation log 5 x = 4 log x 5 Solution: log 5 x = 4 log 5 x. log 5 x = 4 (log 5 x) 2 = 4 log 5 x = 2 or -2 x = 5 2 or 5  2 (Change base from x to 5)