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anonymous

  • 5 years ago

when do you use the inverse when doing logs?

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  1. amistre64
    • 5 years ago
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    log is the inverse of exponents; and exponents are the inverse of logs. What is your question regarding?

  2. anonymous
    • 5 years ago
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    i dont understand how to do them at all. how would you graph y=3log(base5)x. how would you go about that

  3. anonymous
    • 5 years ago
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    Hey amistre can you please help me with 2 antiderivative questions after you are done here?

  4. amistre64
    • 5 years ago
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    Since y = 3 log5(x) is the same as: y = log5(x^3) we can more easily graph its inverse and then "flip" the graph about the y=x line.

  5. amistre64
    • 5 years ago
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    sure thing...

  6. amistre64
    • 5 years ago
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    you got a question posted we can go to :)

  7. anonymous
    • 5 years ago
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    did you get the second equation from switching the x and ys and solving for y?

  8. anonymous
    • 5 years ago
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    did you see my question

  9. amistre64
    • 5 years ago
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    yeah, but I messed it up in my head the first time :)

  10. anonymous
    • 5 years ago
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    meet me there when you can help me

  11. amistre64
    • 5 years ago
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    will do...

  12. amistre64
    • 5 years ago
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    mini: log graphs can be hard to do without a calulator; so we can rewrite it to a more familiar form... do you agree?

  13. anonymous
    • 5 years ago
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    but how do you make it in the other form?

  14. anonymous
    • 5 years ago
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    im there

  15. amistre64
    • 5 years ago
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    y = 3 log5(x) ; divide by 3 y/3 = log5(x) ; take the 5^ of each side 5^(y/3) = 5^(log5(x)) ; 5^(log5) cancel each other out. 5^(y/3) = x Do you agree? Are you familiar with the rules for logs?

  16. anonymous
    • 5 years ago
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    i understand how you did that. when is that that you go about switching the x and y to solve?

  17. amistre64
    • 5 years ago
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    When it makes the graphing easier you can modify it. All you are doing is solving for x instead of y, so keep aware of that

  18. amistre64
    • 5 years ago
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    Would you agree that 5^(y/3) is easier to plot for and solve than log5(x) ? :)

  19. anonymous
    • 5 years ago
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    yess

  20. anonymous
    • 5 years ago
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    how do you do for example logbase8 4096=4

  21. amistre64
    • 5 years ago
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    Do you mean: log8(4096) = 4 ??

  22. anonymous
    • 5 years ago
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    yes

  23. anonymous
    • 5 years ago
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    is it just 8^4=4096

  24. amistre64
    • 5 years ago
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    that is what is known as an identity. one side equals the other. Lets take for example: logB(x) = y this means that B^y=x We can take your equation for instance: log8(4096) = 4 means: 8^4 = 4096, we can test that by either pen and paper , or calculator :)

  25. anonymous
    • 5 years ago
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    35^log 35^x

  26. amistre64
    • 5 years ago
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    35^log35 = 1 and we are left with "x"

  27. anonymous
    • 5 years ago
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    can you do the inverse of y=log1/4 x out step by step please?

  28. anonymous
    • 5 years ago
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    y=ln 6x

  29. amistre64
    • 5 years ago
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    is that log base (1/4)?

  30. anonymous
    • 5 years ago
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    i understand that one now but how do you do the second one?

  31. amistre64
    • 5 years ago
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    y = ln(6x) correct?

  32. anonymous
    • 5 years ago
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    just y= ln 6x not base 6x

  33. amistre64
    • 5 years ago
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    "ln" is just a special way they write log to the base "e"

  34. amistre64
    • 5 years ago
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    y = ln(6x) e^y = e^ln(6x) e^y = 6x (e^y)/6 = x

  35. anonymous
    • 5 years ago
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    how about y= ln (x+2)

  36. amistre64
    • 5 years ago
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    e^y = x+2 e^y -2 = x

  37. anonymous
    • 5 years ago
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    so for the graph of y=log8 x-2 you would do 8^y=x-2 and then fill in values for y such as 0 which would be 1=x-2 x=3?

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