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Math Help - Another confusing Log problems help!

  1. #1
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    Another confusing Log problems help!

    Solve for x
    1.) 4^(2x) - 4^(x+1) - 21 = 0

    2.) 4log*base10*(3x - 2) - log*base10*(x+1) = 1 + log*base10*2
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  2. #2
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    first problem only

    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Solve for x
    1.) 4^(2x) - 4^(x+1) - 21 = 0
    ...
    Hello,

    rearrange your equation to:

    (4^x)^2 - 4*4^x - 21 = 0

    That's a quadratic equation in 4^x. So substitute t = 4^x and you'll get:

    t^2 - 4t - 21 = 0

    Solve this equation for t. (I leave this part for you. There is only one possible solution: 7)

    Re-substitute: 7 = 4^x.
    Now solve this equation for x = (log(10)7) / (log(10)4) approximately 1.4036...

    Greetings

    EB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    2.) 4log*base10*(3x - 2) - log*base10*(x+1) = 1 + log*base10*2
    Let log denote log_10 (log to the base 10). Then the equation is:

    4log(3x - 2) - log(x+1) = 1 + log(2)

    Now the first term may be simplified using A log(B)=log(B^A), so

    log((3x - 2)^4) - log(x+1) = 1 + log(2),

    now using the basic -log(A)=log(1/A) so:

    log((3x - 2)^4) + log(1/(x+1)) = 1 + log(2),

    and now the addition property of logs log(A)+log(B)=log(AB):

    log((3x - 2)^4/(x+1)) = 1 + log(2),

    Now log(10)=1 so the right hand side becomes:

    log((3x - 2)^4/(x+1)) = log(10)+log(2)=log(20).

    Hence:

    (3x - 2)^4 / (x+1) = 20.,

    which is a quartic, which you must solve using whatever techniques you
    have been taught for solving such.

    RonL
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