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anonymous

  • one year ago

Solve 3^(2x) = 7^(x−1).

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  1. anonymous
    • one year ago
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    hellpp

  2. jdoe0001
    • one year ago
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    \(\bf 3^{2x} = 7^{x-1}\) right? for I don't see it happening

  3. anonymous
    • one year ago
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    yes @jdoe0001

  4. jdoe0001
    • one year ago
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    hmmm

  5. jdoe0001
    • one year ago
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    hmm hold the mayo

  6. anonymous
    • one year ago
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    kk

  7. anonymous
    • one year ago
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    @Michele_Laino

  8. jdoe0001
    • one year ago
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    hmmm I assume you've covered logarithms?

  9. anonymous
    • one year ago
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    First step: Take the logarithm of both sides.

  10. anonymous
    • one year ago
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    ao log 3^(2x)= log 7^(x-1)

  11. anonymous
    • one year ago
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    hello?

  12. anonymous
    • one year ago
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    OK. When you have a power, the way to take the log is as follows:\[\log a ^{b} = b \log a\]For example\[\log 5^\left( x+2 \right) = \left( x+2 \right) \log 5\] Try this with your question

  13. anonymous
    • one year ago
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    3^(2x) log= 7^(x-1)= log

  14. anonymous
    • one year ago
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    \[3^{2x}=7^{x-1}=7^x*7^{-1}\] \[\left( 3^2 \right)^x=\frac{ 7^x }{ 7 }\] \[\frac{ 9^x }{ 7^x }=\frac{ 1 }{ 7 }\] \[\left( \frac{ 9 }{ 7 } \right)^x=\frac{ 1 }{ 7 }\] take log and find x

  15. anonymous
    • one year ago
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    −7.74293 7.74293 −1 1 These are the answer choices <3

  16. anonymous
    • one year ago
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    Not quite. Let's look at just the left hand side. You are trying to determine\[\log 3^{2x} = ?\]Compare that with the general case\[\log a^b = b \log a\]Substitute a = 3 and b = 2x. What do you get?

  17. anonymous
    • one year ago
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    2x log 3

  18. anonymous
    • one year ago
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    was i right

  19. anonymous
    • one year ago
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    Correct. Now do the same thing on the right hand side. You are trying to determine\[\log 7^{x-1} = ?\]Applying the same rule, what do you get?

  20. anonymous
    • one year ago
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    x-1=log 7

  21. anonymous
    • one year ago
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    yes?

  22. anonymous
    • one year ago
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    Not exactly. In this one, a = 7 and b = x-1. Try again

  23. anonymous
    • one year ago
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    7 log x-1

  24. anonymous
    • one year ago
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    Backwards. Try again

  25. anonymous
    • one year ago
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    x-1 log 7??

  26. anonymous
    • one year ago
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    i got -7.74293 as an answer

  27. anonymous
    • one year ago
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    That's it. So now you done\[\log 3^{2x} = \log 7^{x-1}\]\[2x \log 3 = \left( x-1 \right) \log 7\]OK so far?

  28. anonymous
    • one year ago
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    yes! I am good

  29. anonymous
    • one year ago
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    Great. Now expand the right hand side.

  30. anonymous
    • one year ago
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    0.47712125472(2x)

  31. anonymous
    • one year ago
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    That's the left hand side. You can multiply the 0.477... by the 2.

  32. anonymous
    • one year ago
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    0.95424250943 sorry

  33. anonymous
    • one year ago
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    Excellent. The left hand side is 0.95424250944 x. Now, on to the left hand side. First thing to do is to expand it.

  34. anonymous
    • one year ago
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    ** right hand side. Sorry

  35. anonymous
    • one year ago
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    0.84509804001x-0.845098804001

  36. anonymous
    • one year ago
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    right??

  37. anonymous
    • one year ago
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    Exactly well done! So now you have\[0.95424250944 x = 0.84509804001 x - 0.8450904001\]Can you gather up the x's on one side and solve?

  38. anonymous
    • one year ago
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    -0.36797678529=-0.8450904001

  39. anonymous
    • one year ago
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    so would the answer be 7.74293?

  40. anonymous
    • one year ago
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    Problem with the left hand side. Remember, you are subtracting 0.84509804001 x from both sides. Try the left hand side again.

  41. anonymous
    • one year ago
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    0.10914370543x

  42. anonymous
    • one year ago
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    That's better. So you have\[0.10914370543 x = 0.84509804001\]To solve for x, divide both sides by 0.10914370543. What do you get?

  43. anonymous
    • one year ago
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    7.74298468868 !!!!

  44. anonymous
    • one year ago
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    Yayyyy!! Well done!

  45. anonymous
    • one year ago
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    Yayyyy! If I post another question in the open section, will u answer?

  46. anonymous
    • one year ago
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    thanks:)

  47. anonymous
    • one year ago
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    Yup. You're welcome

  48. anonymous
    • one year ago
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    To avoid dealing with all those decimal places, most folks will leave the logs until the end, for example\[2x \log 3 = (x-1)\log 7\]\[(2 \log 3) x = (\log 7) x - \log 7\]\[(2 \log 3 - \log 7) x = -\log 7\]\[x = \frac{ -\log 7 }{ 2 \log 3 - \log 7 }\]Then plug it into your calculator

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