1 / 21

Exponential and Logarithmic Equations (4.6)

Exponential and Logarithmic Equations (4.6). The Change of Base Formula and Using various patterns to solve equations. Formative Assessment Review. I heard from several of you that going over how to find an inverse function would be helpful. Here we go:. Formative Assessment Review.

kasimir-moon
Download Presentation

Exponential and Logarithmic Equations (4.6)

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. Exponential and Logarithmic Equations (4.6) The Change of Base Formula and Using various patterns to solve equations

  2. Formative Assessment Review I heard from several of you that going over how to find an inverse function would be helpful. Here we go:

  3. Formative Assessment Review The graphs: The last one on page 3, with variables: Also, on set four (the final page), be sure to LABEL! Without some indication of intervals, the graphs are just scribbles. And BE NEAT! I expect to see curves that line up along asymptotes, and cross at actual intercepts.

  4. Formative Assessment Review Final hint– review your notes regularly! Even five minutes a few times a week makes a big difference.

  5. Using properties from last time Solve the equation: ln(x+6) - ln10 = ln (x-1) - ln2

  6. Using properties from last time Solve the equation: ln(x+6) - ln10 = ln (x-1) - ln2 ln ((x+6)/10) = ln ((x-1)/2) (x+6)/10 = (x-1)/2 2(x+6) = 10(x-1) 2x +12 = 10x -10 22 = 8x x = 11/4

  7. Using properties from last time In economics, the demand D for a product is often related to its selling price p by an equation of the form logaD = logac - k logap Where a and c are positive constants (why positive?) and k > 1. • Solve the equation for D. • How does increasing or decreasing the selling price affect the demand? • Think about a and c.

  8. Using properties from last time In economics, the demand D for a product is often related to its selling price p by an equation of the form logaD = logac - k logap Where a and c are positive constants and k > 1. • Solve the equation for D.

  9. Using properties from last time In economics, the demand D for a product is often related to its selling price p by an equation of the form logaD = logac - k logap D = c/pk Where a and c are positive constants and k > 1. • How does increasing or decreasing the selling price affect the demand?

  10. Using properties from last time In economics, the demand D for a product is often related to its selling price p by an equation of the form logaD = logac - k logap Where a and c are positive constants and k > 1. • Think about a and c. What do you think a would equal? In other words, what would be a reasonable log base? What do you think c represents? (Econ students, help here.)

  11. Now, a logarithmic POD Rewrite for x: 3x = 21 What is a reasonable guess for the value of x?

  12. From the POD How to solve for x? 3x = 21 x = log321 Start by taking the log of each side. log 3x = log 21 Then use our third Log Property. x log 3 = log 21 Then solve for x. x = (log 21)/(log 3) x = 2.77 If I came up with these values, what have I done? x = .845 x = 7 x = 1.64

  13. From the POD x = log321 becomes x = (log 21)/(log 3) This leads to the Change of Base Formula: logb u = (loga u)/(loga b) = (log u)/(log b) = (ln u)/(ln b)

  14. The Proof (The POD with Variables) Set w = logb u. Then bw = u Log each side loga bw = loga u w loga b = loga u w = (loga u)/(loga b) = logb u

  15. Careful! loga (u/b) = loga u - loga b It does not equal (loga u)/(loga b). (Major foot stomp here.)

  16. Use it • Solve for x when there are different bases. 42x+3 = 5x-2 Start by taking the log of each side. That way you set up a common base. Then you can move the exponents down. log 42x+3 = log 5x-2 (2x+3)log 4 = (x-2)log 5 And solve.

  17. Use it • Solve for x when there are different bases. 42x+3 = 5x-2 log 42x+3 = log 5x-2 (2x+3)log 4 = (x-2)log 5 .602(2x+3) = .699(x-2) 1.204x + 1.806 = .699x – 1.398 .505x = -3.204 x = -6.345 Check it!

  18. Use it • Solve for x. (Alert! Embedded quadratic here.) (5x - 5-x)/2 = 3

  19. Use it • Solve for x. (Alert! Embedded quadratic here.) Does it check?

  20. Use it • If a beam of light with intensity I0 is projected vertically downward into a certain body of water, then its intensity, I(x), at a depth of x meters is I(x) = I0e-1.4x At what depth is its intensity half of its value at the surface?

  21. Use it • If a beam of light with intensity I0 is projected vertically downward into a certain body of water, then its intensity, I(x), at a depth of x meters is I(x) = I0e-1.4x So, the intensity is cut in half at about half a meter.

More Related