Question

Hi all! This one has been troubling me. Solve the inequality: av(x-1)-av(x-3) >or equal to 5

Answer 1

\(\large a^{x-1} - a^{x-3} \ge 5\)

Answer 2

is it like above ?

Answer 3

umm let me try to make the equation

Answer 4

okie

Answer 5

\[\left| x-1 \right|-\left| x-3 \right| \ge 5\]

Answer 6

it just keeps cancelling out when I do it, so I want to be sure I'm not doing something wrong

Answer 7

okay ! its a combination of two absolute value functions

Answer 8

at wat point u got stuck ?

Answer 9

ok

Answer 10

so I know it can be positive/negative, but when when I solve I get an answer that doesnt make sense like 2 is greater than/equal to 5

Answer 11

good :)
that means, there are no solutions for this inequality

Answer 12

wat u arrived at is correct, there are no solutions for this guy

Answer 13

aha, I just get paranoid when there isn't a solution XD

Answer 14

lol who doesnt !
btw how did u solve it ?

Answer 15

umm well I had two solutions

Answer 16

did u square both sides ?

Answer 17

?

Answer 18

I erased it let me rewrite real quick

Answer 19

its okay :)
just wondering wat method u used.

i did this by squaring both sides, to get rid of absolute bars

Answer 20

If x>0 2>or equal to 5 for example

Answer 21

then for x<0

Answer 22

what is the squaring method?

Answer 23

\(\left| x-1 \right|-\left| x-3 \right| \ge 5 \)

\(\left| x-1 \right|~~~\ge~~~ 5 +\left| x-3 \right| \)

Answer 24

square both sides

Answer 25

\(\left| x-1 \right|-\left| x-3 \right| \ge 5 \)

\(\left| x-1 \right|~~~\ge~~~ 5 +\left| x-3 \right| \)

\((x-1)^2~~~\ge~~~ 25 + (x-3)^2 + 10\left| x-3 \right| \)

Answer 26

I see interesting

Answer 27

if u simplify and keep going,
u wil arrive at 2=3 or coconuts = oranges kindof statement
so we conclude NO solution exist

Answer 28

cool! I didn't know you could do that

Answer 29

Thanks again for you help as usual :))

Answer 30

\(\left| x-1 \right|-\left| x-3 \right| \ge 5 \)

\(\left| x-1 \right|~~~\ge~~~ 5 +\left| x-3 \right| \)

\((x-1)^2~~~\ge~~~ 25 + (x-3)^2 + 10\left| x-3 \right| \)

....

\( 4x-33 ~~~\ge~~~ 10\left| x-3 \right| \)

square both sides again

....

Answer 31

np :)

Answer 32

^_^