# 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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