Question

2^(a-b)=6
2^(a+b)=12 find b=?

Answer 1

well just try logarithm?

Answer 2

log2,6=a-b
log 2,12=a+b

Answer 3

wait i take my notebook and my pencil ^^

Answer 4

Can we do it like this:
2^(a-b)=6
-----------=2^(a-b-a-b)=1/2
2^(a+b)=12

2^-2b=1/2
2^-2b=2^-1
-2b=-1
b=1/2

Answer 5

I don't think we need log....

Answer 6

m2 just its a way ...

Answer 7

you can do that since the bottom w know dont equal 0

Answer 8

kk thnx ^_^
@Amistre can u pls look 1 of my q?

Answer 9

which one?

Answer 10

its simple
2^a - 2^b = 6
2^a + 2^b=12

Answer 11

this 1:
What should m be so that :
(m+3)x^2-5mx+4m+1>0 for every x...

Answer 12

and now its 2 x 2^a = 18 ^^

Answer 13

Korcan 2^(a-b)=2^a/2^b

Answer 14

(m+3)x^2 +(-5m) x + (4m +1) > 0 its a quad that needs to be accounted for a b and c

Answer 15

\[x = 0; when;x=\frac{5m \pm \sqrt{25m^2 -4(m+3)(4m+1)}}{2(m+3)}\]

Answer 16

we can narrow down the results by working the discriminate

Answer 17

yea i found this 2..but i can't go any further...O.o

Answer 18

u see u are better then me...

Answer 19

right..:P same story :P

Answer 20

25m^2 -4m^2 -16m -12 >= 0

Answer 21

21m^2 -16m -12 >= 0 ; quad form that to see your options

Answer 22

tht's it? Thnx ^_^ :)

Answer 23

well angela can you help me on my question?

Answer 24

hmmmm..idk..let me see

Answer 25

the roots to that are:

m= \[\frac{2}{21}(4\pm \sqrt{79})\]

Answer 26

...... i missques the discriminate when I did that :/

Answer 27

Sorry but i need some english lessons too..wht's missques? O.o