Asked by exponential growth
If log_P(Q) = x^2 and log_Q(P)=1/(2x-1) find x.
Answers
Answered by
oobleck
log_p(q) = 1/log_q(p), so
x^2 = 2x-1
x = 1
x^2 = 2x-1
x = 1
Answered by
mathhelper
did you mean:
log<sub>P</sub>Q = x^2 and log<sub>Q</sub> P = 1/(2x-1) find x. ?????
I will assume you do ...
A lesser know property of logs is that
if log<sub>a</sub>b = x
then log<sub>b/sub>a = 1/x
that is:
log<sub>a</sub>b = 1/log<sub>b/sub>a
e.g. check on your calculator that log<sub>5/sub>3 = 1/log<sub>3/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
log<sub>P</sub>Q = x^2 and log<sub>Q</sub> P = 1/(2x-1) find x. ?????
I will assume you do ...
A lesser know property of logs is that
if log<sub>a</sub>b = x
then log<sub>b/sub>a = 1/x
that is:
log<sub>a</sub>b = 1/log<sub>b/sub>a
e.g. check on your calculator that log<sub>5/sub>3 = 1/log<sub>3/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
Answered by
mathhelper
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
good to know it works, just forgot to close the one loop
Answered by
oobleck
I always copy/paste the <sub>?</sub> pairs, then fill in the contents.
I've been stung before, too!
I've been stung before, too!
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