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acc. alg 2

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exponentiating to solve equations: solve the eaquation, cheack for extraneous solutions

log5(x+4)+log5(x+1)=2

  • acc. alg 2 - ,

    remember log a + log b= log ab

    do that, then take the antilog..

    log5 (x+4)(x+1)=2

    (x+4)(x+1)=25
    then proceed to solve.

  • acc. alg 2 - ,

    log5 [(x+4)(x+1)] = 2

    5^log5 [(x+4)(x+1)] = 5^2 = 25

    (x+4)(x+1) = 25

    x^2 + 5 x + 4 = 25

    x^2 + 5 x - 21 = 0

    x = [-5 +/- sqrt(25+84) ]/2

    = [-5 +/- sqrt(109)]/2

    = [ -2.5 +/- 5.22 ]
    = 2.72 or - 7.72
    You can not take log of a negative number so use 2.72

  • acc. alg 2 - ,

    i dnt get bwat u did after x^2+5x-21=0. that is were i keep gettin stuck on

  • acc. alg 2 - ,

    http://en.wikipedia.org/wiki/Quadratic_equation

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