Question

Can anyone tell me the first step on solving this?

log1/2 (x-1) - log1/2 (x+1) < log1/2 (x-2) + 1

Answer 1

first step is to try to combine in a single log, so you can write in equivalent exponential form

Answer 2

lol they all have the same base so i can equal, right? i forgot that

Answer 3

(x-1) - (x+1) < (x-2) + 1
can i do this?

Answer 4

All I got was x<x^2

Answer 5

lol i got x>-1 and with the conditions i got x>2 but thats not in my options

Answer 6

thx anyways mads <3

Answer 7

@KingGeorge any ideas?

Answer 8

Can you clarify what the expression is in the equation editor?

Answer 9

1/2 is the base.
Ok, I will try

Answer 10

1/2 is the base. Then, while you're idea is a good one, we can't do it yet because there's a term that isn't in a the same log function yet. Namely, the term 1.

Answer 11

I thought of that. May I use the conversion of bases formula with that one?

Answer 12

However, to put 1 in a function, note that \(log_{1\over2}(x)=1 \longrightarrow x= {1\over2}\) So instead of 1, use \[log_{1\over2}\left({1\over2}\right)\]Then you can take away the log functions because they're all equal.

Answer 13

x= 3 or -1

Answer 14

I redid the math on paper..

Answer 15

\[\log _{1/2} (x-1) - \log _{1/2} (x+1) < \log _{1/2} (x-2) + 1\]

Answer 16

oh wait greater than sign...

Answer 17

So \((x-1)-(x+1)<(x-2)+1/2\) so \(-2 < x-3/2\) so we finally get that \[x>-{1\over2}\]

Answer 18

That's weird... I will post my options

Answer 19

I made a mistake with my second calculation too lol!

Answer 20

Nevermind, I made a mistake too. You can only remove the logs if the entire side is contained within a single logarithm. So instead, put the log around the 1, combine the logs, and then remove them.

Answer 21

a) \[\left\{ x R / 2 \le x \le 3\right\}\]
b) \[\left\{ x \in R /2<x <3 \right\}\]
c) \[\left\{ xR/0<x <3\right\}\]
d) \[\left\{ xR/0 \le x \le3 \right\}\]