If log_P(Q) = x^2 and log_Q(P)=1/(2x-1) find x.

Answer 1

log_p(q) = 1/log_q(p), so
x^2 = 2x-1
x = 1

Answer 2

did you mean:
log_PQ = x^2 and log_Q P = 1/(2x-1) find x. ?????
I will assume you do ...

A lesser know property of logs is that
if log_ab = x
then log<sub>b/sub>a = 1/x
that is:
logab = 1/logb/sub>a

e.g. check on your calculator that log5/sub>3 = 1/log3/sub>5

from yours ....
x^2 = 1/[1/(2x - 1) ] = 2x - 1
x^2 - 2x + 1 = 0
(x-1)^2 = 0
x-1 = 0
x = 1</sub>

Answer 3

argghhhh, that's what you get by trying to be fancy with code.

good to know it works, just forgot to close the one loop

Answer 4

I always copy/paste the _? pairs, then fill in the contents.
I've been stung before, too!