1. profile image44
    nicole6posted 8 years ago


    solve for x

  2. nadp profile image78
    nadpposted 8 years ago

    I'm assuming that the exponents are 5x-1 and 3(x+11) even though there are nor parentheses around this expressions.
    If that is the case then you would take the common logarithm of both sides:

    log(4^5x-1) = log(2^3(x+11))

    Then you could use the power rule for logs to get:

    (5x-1)(log4) = (3(x+11))(log2)

    simplify, and get variable terms on the same side:

    (5x-1)/(3x+33) = (log2)/(log4)

    use FOIL to multiply the left side, and a calculator to find the value of the right side:

    15x^2+162x-33 = .5

    subtract .5 from both sides:

    15x^2+162x-32.5 = 0

    use the quadratic formula to solve for x:

    xsad(-162)+or-sqr rt(162^2-4(15)(-32.5))/(2(15))

    xsad-162+or-sqr rt(28194))/30


    x = -131.34 or -192.66

    where'd this problem come from???