Question

(1/3)^8x-1=27

Answer 1

27=(1/3)^(-3)

Answer 2

how?

Answer 3

http://usablealgebra.landmark.edu/algebra/exponents/rules.php
(1/3)^(-3) = ((1/3)^(-1))^3 = 3^3 = 27

Answer 4

once the bases are equal, set the exponents equal:
8x-1 = -3

Answer 5

this method works for all equations that are one to one logs?

Answer 6

yup. you can also you logs for this, but there is no need since the bases can be made the same

Answer 7

its precal so i have to use logs

Answer 8

using logs: log( (1/3)^(8x-1) ) = log(27)
(8x-1) log(1/3) = log(27)
(8x-1)*(-log(3)) = log(27)
8x-1= - log(27)/log(3)
8x-1 = - log base 3 of 27
8x-1 = -3

log(a)/log(b) = log base b of a