Algebra II
posted by Lucy .
Solve:
(1/81)^t=243^t2
The book states to get the same base but I don't understand.
Thanks for any help.

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^t2
(3^4)^t = (3^5)^(t2)
3^(4t) = 3^(5t10)
then 4t = 5t  10
9t=10
t=10/9
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