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JayDS

  • 2 years ago

Find the equation of the inverse of the following.

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  1. JayDS
    • 2 years ago
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    \[y=3^{x}\] \[y=_{5}x\]

  2. Jusaquikie
    • 2 years ago
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    the inverse is just if you put in an x and get a y the inverse is like putting in the y and getting the same x you put in the first part.

  3. JayDS
    • 2 years ago
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    yep, I did that.

  4. Jusaquikie
    • 2 years ago
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    what you can do is switch your x's and y's then just solve for y.

  5. JayDS
    • 2 years ago
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    \[y=3^{x}\] \[x=3^{y}\] \[\log _{10}x = \log _{10}3^{y}\] \[\log _{10} x = ylog _{10}3\] \[y=\frac{\log _{10}x}{\log _{10}3}\] y=x/3?

  6. JayDS
    • 2 years ago
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    I have answers but I want to know what I did wrong, because it seems right to me.

  7. Jusaquikie
    • 2 years ago
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    i think you need to keep the logs or put e^y

  8. Jusaquikie
    • 2 years ago
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    you can only take away the logs like you did if y had one on that side

  9. Jusaquikie
    • 2 years ago
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    i think it can be rewrote as y=ln(x-3) but i'm a little rusty with logarithms

  10. JayDS
    • 2 years ago
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    but there is log10 on both sides, so I can cancel it out? and ok... I'm still confused sorry.

  11. Jusaquikie
    • 2 years ago
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    when you moved the y over to the other sice you had y=log(x)/log(3) y doesn't have a log

  12. Jusaquikie
    • 2 years ago
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    if it was log(y)=log(x)/log(3) you could cancel them out

  13. JayDS
    • 2 years ago
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    really? hmmm, confusing but I know what you mean.

  14. Jusaquikie
    • 2 years ago
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    yeah i get confused too when you start using logs

  15. Jusaquikie
    • 2 years ago
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    if you have the answer try to work it backwards to get the first question and try to tie the two to gether with what your sure of.

  16. JayDS
    • 2 years ago
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    the answer is \[\log _{3}x for x>0\] can you help me out? I don't understand how they got it.

  17. Jusaquikie
    • 2 years ago
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    log 2 x or log 3 x?

  18. Jusaquikie
    • 2 years ago
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    \[\log_{a}y=x \] is \[a^{x}=y\]

  19. JayDS
    • 2 years ago
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    3x

  20. JayDS
    • 2 years ago
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    from log10(x)=ylog10(3) can you show me the rest of the steps?

  21. Jusaquikie
    • 2 years ago
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    so we can rewriteyour first one as as\[\log_{3} x=y\]

  22. JayDS
    • 2 years ago
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    after you switch the x and y I believe?

  23. Jusaquikie
    • 2 years ago
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    sorry i'm trying to remember this as we go

  24. JayDS
    • 2 years ago
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    it's ok.

  25. Jusaquikie
    • 2 years ago
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    no that is just rewriting it then we swith and solve for x

  26. JayDS
    • 2 years ago
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    you wrote it wrong then.

  27. Jusaquikie
    • 2 years ago
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    all i did was change the exponent to a log

  28. JayDS
    • 2 years ago
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    \[y=3^{x}\] \[\log _{3}y=x\]

  29. JayDS
    • 2 years ago
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    I think you got confused with the x and y, you just got them in the wrong position.

  30. JayDS
    • 2 years ago
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    then we can switch the x & y from there.

  31. Jusaquikie
    • 2 years ago
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    it's hard to do math on these boards lol

  32. JayDS
    • 2 years ago
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    yeh I know haha, especially trying to write the equation takes years.

  33. JayDS
    • 2 years ago
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    using the equation tab*

  34. Jusaquikie
    • 2 years ago
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    i think i have it let me check with my calculator

  35. JayDS
    • 2 years ago
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    kk.

  36. Jusaquikie
    • 2 years ago
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    ok lol i can't figure out how to do it on my calculator but i think it's right, so all we are doing is changing the question into a log of the same equasion then when we swap x and y we already have the answer

  37. Jusaquikie
    • 2 years ago
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    by the rules of logarithms we have loga^b=c is the same as a^c=b

  38. JayDS
    • 2 years ago
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    ok, and it turns out to be\[y=\log _{3}x\] right?

  39. Jusaquikie
    • 2 years ago
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    so taking your problem y=3^x and putting it in log form we get log3^y=x then we swap y and x to get log3^x=y is that right?

  40. JayDS
    • 2 years ago
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    yes.

  41. Jusaquikie
    • 2 years ago
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    so let's try the second one

  42. Jusaquikie
    • 2 years ago
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    y=5^x is log5^y=x so swap is log5^x=y

  43. JayDS
    • 2 years ago
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    yep but the answer only says 5^x for some reason.

  44. JayDS
    • 2 years ago
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    confusing, I'll just have to ask my teacher. I'm just doing it advanced anyway.

  45. JayDS
    • 2 years ago
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    thanks for the help.

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