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anonymous

  • one year ago

Solve 4^(2x) = 7^(x−1).

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  1. anonymous
    • one year ago
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    @Astrophysics @rishavraj @ganeshie8

  2. freckles
    • one year ago
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    have you tried taking ln( ) of both sides

  3. anonymous
    • one year ago
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    wouldn't it be log?

  4. freckles
    • one year ago
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    \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7)\] you can use log too if you want

  5. freckles
    • one year ago
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    not distribute on the right hand side

  6. freckles
    • one year ago
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    and put your terms with x in it on one side and your terms without x on the opposing side

  7. anonymous
    • one year ago
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    1.38629436112(2x) I mean, sorry :)

  8. freckles
    • one year ago
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    are you allowed to approximate ?

  9. anonymous
    • one year ago
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    no, not until we have the final answer

  10. freckles
    • one year ago
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    ok so leave that one side as 2ln(4)x and distribute on the other side recall the distributive property is a(b+c)=ab+ac

  11. freckles
    • one year ago
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    use distributive property here and what do you get: \[\ln(7)(x-1)=?\]

  12. anonymous
    • one year ago
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    1.94591014906

  13. anonymous
    • one year ago
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    1.94591014906x-1.94591014906

  14. freckles
    • one year ago
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    I think you mean to say ln(7)x-ln(7)

  15. anonymous
    • one year ago
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    oh yes! Sorry

  16. freckles
    • one year ago
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    \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \]

  17. freckles
    • one year ago
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    now to get the x people together subtract ln(7)x on both sides

  18. freckles
    • one year ago
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    \[ax=bx-c \\ ax-bx=-c \\ x(a-b)=-c \\ \text{ the last step is to choose to divide by } \\ \text{ what is in front of the } x \\ \\ \text{ in this example that would be } (a-b) \\ x=\frac{-c}{a-b}\]

  19. freckles
    • one year ago
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    this is exactly what you are going to do here first subtract ln(7)x on both sides then use my example as a guide sorta to finally find x

  20. anonymous
    • one year ago
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    2.77258872224x=1.94591014906x-1.94591014906

  21. freckles
    • one year ago
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    I thought you aren't allowed to approximate until you have final answer

  22. anonymous
    • one year ago
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    i didn't :)

  23. freckles
    • one year ago
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    ln(7) is irrational there is no way you can write out the whole number same for ln(4)

  24. anonymous
    • one year ago
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    oh. Sorry, I just typed it into my calculator & thats what it said

  25. freckles
    • one year ago
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    your calculator can only show you so many digits

  26. freckles
    • one year ago
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    it doesn't have an infinite screen display

  27. freckles
    • one year ago
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    anyways... \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \\ \text{ the step I was asking you \to do is subtract } \ln(7) x \text{ on both sides } \\ 2 \ln(4) x -\ln(7) x=-\ln(7)\] try factoring the x out on the left hand side like I did in my example ax-bx=x(a-b)

  28. anonymous
    • one year ago
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    2 In(x)-In(-3)x=-In (7)

  29. freckles
    • one year ago
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    |dw:1438798478614:dw|

  30. freckles
    • one year ago
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    \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \\ \text{ the step I was asking you \to do is subtract } \ln(7) x \text{ on both sides } \\ 2 \ln(4) x -\ln(7) x=-\ln(7) \\ ( 2 \ln(4)-\ln(7) )x=-\ln(7)\] I'm going to leave the last step to you

  31. anonymous
    • one year ago
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    Alright, so I got 2 In (-3) =-In(7) In(-3)^2=-In(7) yes?

  32. anonymous
    • one year ago
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    2.35389<---is my answer

  33. anonymous
    • one year ago
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    yes?

  34. anonymous
    • one year ago
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    I gotta go eat lunch:) Please message me if I was wrong

  35. freckles
    • one year ago
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    not sure where you got 2 ln(-3)=-ln(7) just divide both sides by what x is being multiplied by which is 2ln(4)-ln(7) this does not equal 2 ln(-3) divide both sides by 2ln(4)-ln(7)

  36. freckles
    • one year ago
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    \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \\ \text{ the step I was asking you \to do is subtract } \ln(7) x \text{ on both sides } \\ 2 \ln(4) x -\ln(7) x=-\ln(7) \\ ( 2 \ln(4)-\ln(7) )x=-\ln(7) \\ x=\frac{-\ln(7)}{2 \ln(4)-\ln(7)}\]

  37. anonymous
    • one year ago
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    -3.29?

  38. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=-ln%287%29%2F%282ln%284%29-ln%287%29%29 this calculator says that is approximately -2.35

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