Math, Precalc

posted by Susan

(3^x)+(5^{x+3})=(3^{x+4})-(1/3)*( 5^{x+2}).

Solve

  1. Anonymous

    3^x + 5^(x+3) - 3^(x+4) + (1/3)5^(x+2)= 0
    3^x(1 - 3^4) + 5^(x+2)(5 + (1/3)(1) ) = 0
    -80(3^x) + (16/3)(5^(x+2) = 0
    times -3/16
    15(3^x) - 5^(x+2) = 0
    15(3^x) = 5^(x+2)
    divide by 5
    3(3^x) = 5^(x+1)
    3^(x+1) = 5^(x+1)
    this can only be true if x = -1
    that is, if 3^0 = 5^0 , which is true

  2. Reiny

    3^x + 5^(x+3) - 3^(x+4) + (1/3)5^(x+2)= 0
    3^x(1 - 3^4) + 5^(x+2)(5 + (1/3)(1) ) = 0
    -80(3^x) + (16/3)(5^(x+2) = 0
    times -3/16
    15(3^x) - 5^(x+2) = 0
    15(3^x) = 5^(x+2)
    divide by 5
    3(3^x) = 5^(x+1)
    3^(x+1) = 5^(x+1)
    this can only be true if x = -1
    that is, if 3^0 = 5^0 , which is true

  3. Steve

    just start by combining like powers:

    (3^x)+(5^{x+3})=(3^{x+4})-(1/3)*( 5^{x+2})
    3^x + 125*5^x = 81*3^x - 3^-1 * 25*5^x

    3*3^x + 375*5^x = 243*3^x - 25*5^x
    (375+25)*5^x = (243-3)*3^x
    400*5^x = 240*3^x
    5*5^x = 3*3^x
    5^(x+1) = 3^(x+1)
    (x+1)log5 = (x+1)log3
    (x+1)(log5-log3) = 0
    x = -1

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