can someone pleaaase help me with these "solve for x":
a) log_x 100 = 5
b) 3^x+5 = 81
c) log (4x + 2) - log 2 = log 6x
thanks heaps!
(a) x^5 = 100, so x = 100^(1/5)
(b) I assume you mean
3^(x+5) = 81
we know that 3^4 = 81, so
x+5 = 4
(c) log (4x+2)/2 = log 6x
the logs are equal, so we just have
(4x+2)/2 = 6
I guess you can probably handle that one, eh?