1 / 10

SOLVING LOGARITHMIC AND INDICES PROBLEM

This guide provides step-by-step solutions for equations in the form of ( ax = ay ) and explores various examples of solving logarithmic equations, indices problems, and using logarithms to simplify complex expressions. Key topics include equating indices, solving through logarithms, and handling different bases effectively. We’ll clarify common mistakes and provide corrective examples to enhance understanding. With practical applications and clear illustrations, this resource is invaluable for mastering logarithmic and indices equations.

lang
Télécharger la présentation

SOLVING LOGARITHMIC AND INDICES PROBLEM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related