Algebra II

Solve:

(1/81)^t=243^t-2

The book states to get the same base but I don't understand.

Thanks for any help.

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  1. the key is to recognize
    both 81 and 243 as powers of 3

    81 = 3^4 and 243 = 3^5

    so (1/81)^t=243^t-2
    (3^-4)^t = (3^5)^(t-2)

    3^(-4t) = 3^(5t-10)

    then -4t = 5t - 10
    -9t=-10
    t=10/9

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