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SOLVING LOGARITHMIC AND INDICES PROBLEM

SOLVING LOGARITHMIC AND INDICES PROBLEM. 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. Examples. 8 x+1 = 4 x+3 (2 3 ) x+1 = (2 2 ) x+3 2 3x+3 = 2 2x+6

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SOLVING LOGARITHMIC AND INDICES PROBLEM

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  1. SOLVING LOGARITHMIC AND INDICES PROBLEM

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

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

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

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

  6. SOLVE 2X + 2X+3 = 32 2x + 2x  23 = 32 Factorize 2x 2X(1 + 23) = 32 2X (9)= 32 2X = 32/9 X lg 2 = lg 32 – lg 9 X (0.3010)= 1.5051-0.9542 X=1.8302

  7. INDEX EQUATION WITH DIFFERENT BASE Solving equation in the form of ax = b,where a ≠-1, 0 , 1 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. log10 5 x = log10 6 x log10 5 = log10 6 x (0.6990) = 0.7782 x = 0.7782  0.6690 x = 1.113

  8. Example: Solve 5x – 3x+1 = 0 Solution: 5x – 3x+1 = 0 5x = 3x+1 Taking logarithms on both sides, lg 5x = lg 3x+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 log5 (5x – 4) = 2 log5 3 + log5 4 First, simplify the right hand side. log5 (5x – 4) = log5 3 log5=log5 Comparing number in both sides. log5log5 5x = 40 x = 8 2  log5 4 + (5x – 4 ) (36) (5x – 4) = (36)

  10. Solve the equation log5 x = 4 logx 5 Solution: log5 x = 4 log5 x. log5 x = 4 (log5 x)2 = 4 log5 x = 2 or -2 x = 52 or 5 2 (Change base from x to 5)

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