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Math Help - exponents and logs

  1. #1
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    exponents and logs

    This type of question is kiiiiiling me right now.
    Solve for x.
    2^(1-x) = 3
    Solve for x.
    3^(2x-1) = 5

    EDIT: these two are similar ones I can't find a solid approach to.
    log base 2 of (log base 3 of (x)) = 4
    and
    10^5^x = 3

    I thought I would brush up on my exponents and logs before I did the exponents and logs chapter in my calculus text, but I can't remember how to solve this kind of thing, thank you!
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  2. #2
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    Re;

    Re:
    Attached Thumbnails Attached Thumbnails exponents and logs-11.gif  
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  3. #3
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails exponents and logs-12.gif  
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  4. #4
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    log_2(log_3(x))=4
    log_3(x)=2^4
    x=3^(2^4)
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  5. #5
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    10^5^x=3
    log_10(10^5^x)=log_10(3)
    5^x=log_10(3)
    log_5(5^x)=log_5(log_10(3))
    x=log_5(log_10(3))
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  6. #6
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    Actually there are two possible meanings to 10^5^x:
    (10^5)^x
    and
    10^(5^x)

    They are NOT equal so you need to use parenthesis!!

    -Dan

    Edit: I suppose by order of operations 10^5^x = (10^5)^x, but I would prefer the parenthesis for clarity.
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  7. #7
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    Hello, turillian!

    We can solve these without the base-change formula . . .


    Solve for x: .3^(2x-1) .= .5
    Take logs of both sides. .The base doesn't matter; use ln.

    . . . . . ln(3^{2x-1}) .= .ln(5)

    Then: .(2x - 1)·ln(3) .= .ln(5)

    . . . . . . . . . .2x - 1 .= .ln(5)/ln(3)

    . . . . . . . . . . . . 2x .= .ln(5)/ln(3) + 1

    . . . . . . . . . . . . .x .= .˝[ln(5)/ln(3) + 1]


    Therefore: .x . .1.23248676



    log2[log3(x)] .= .4
    Rewrite in exponential form.

    We have: .log
    3(x) .= .2^4 . . . . log3(x) .= .16

    Rewrite again: .x .= .3^{16}

    Therefore: .x .= .43,046,721




    10^5^x .= .3
    Conventionally, a "stack" of exponents is read from the top.

    We have: .10^(5^x) .= .3

    Rewrite in logarithmic form: .5^x .= .log
    10(3)

    Rewrite again: .x .= .log
    5[log10(3)]

    Therefore: .x . .-0.459778294

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  8. #8
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    Thanks for the help everyone.
    Yes, the stack of exponents was meant to be read from the top. It was
    10^(5^x).
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