pre calc

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find the exact solution algebraically and check it by substituting into the orginal equation.

1. 36(1/3)^x/5=4

2. 32(1/4)^x/3=2

  • pre calc -

    the way you typed it .....
    36(1/3)^x = 20
    (1/3)^x = .5555555..
    x ln(1/3) = .55555..
    x = .535 appr.

    what you probably meant:
    36(1/3)^(x/5) = 4
    (1/3)^(x/5) = .111111..
    x/5 = ln .111111 / ln(1/3)
    x = 10

    in the second, you probably have the same typo, and I will assume:
    32(1/4)^(x/3) = 2

    here is another way to do this:
    notice that all factors are powers of 2, so ...
    (2^5)(2^-2)^(x/3) = 2^1
    2^(-2x/3) = 2^-4

    -2x/3 = -4
    x = 6

    we could have done the first the same way, after dividing both sides by 32 to get
    (1/3)^(x/5) = 1/9 = (1/3)^2
    so x/5 = 2
    x = 10

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