Question

Reviewing and in need of a refreshment.

Answer 1

\[\log_{5}(3y-1)=\log_{5}(y+4)\]

Answer 2

by applying the same function on both sides, you obtain a much simpler expression. see whatt it is?

Answer 3

can't you take 5^{log_5}?

Answer 4

(the one that is the inverse (recirocal) of \(\log_5\))

Answer 5

^

Answer 6

\[\large 5^{\log_{5}(3y-1)}=5^{\log_{5}(y+4)}\]

Answer 7

you can't do that? :;\ or YOU CAN! i believe you can. :)

Answer 8

@jim_thompson5910 jimmm jim jim

Answer 9

since the bases are same on both sides thus we can have :