Question

efreferwtfr

Answer 1

@Mertsj @phi

Answer 2

a=2?
b=(ln(3)-ln(2))/(2ln(2))?

Answer 3

no b is wrong. how would you solve for b

Answer 4

b(0)=0

Answer 5

b=(ln(3)-ln(2))/(2ln(2))?

close... where did the extra 2 come from in the bottom

Answer 6

haha not sure since it's wrong. which two is the extra?

Answer 7

at the point 1,3 (and with a=2)
2 * 2^(b*1) = 3

Answer 8

oh is it just ln(2) as the denominator?

Answer 9

you could do a few things
2^1 * 2^b is 2^(b+1)
so you have
2^(b+1) = 3

solve by taking the log of both sides

or
2* 2^b= 3
divide both sides by 2

2^b= 3/2
solve by taking the log of both sides

Answer 10

yup i got it

Answer 11

can you go through the steps?

Answer 12

yeah you get 2^b=3/2 and then you take the natural log of both sides which makes it b=ln(3/2)/ln(2)

Answer 13

yes,
ln(2^b) is b*ln(2)
ln(3/2) can also be written ln(3) - ln(2)
b= ln(3)/ln(2) - ln(2)/ln(2) = ln(3)/ln(2) - 1

if we did it the other way
2^(b+1) = 3
(b+1) * ln(2) = ln(3) (after taking the log of both sides
divide by ln(2):

b+1 = ln(3)/ln(2)
sub 1
b= ln(3)/ln(2) -1

Answer 14

@habibmatatta
Now, as I was saying...
2=a(2)^b(0)
2=a(2)^0
2=a(1)
a=2
Also from the second ordered pair
3=a(2)^(b(1)
3=2(2)^b
1.5=2^b
log 1.5=b(log 2)
.58=b
y=2(2)^.(58x)

Answer 15

ok thanks

Answer 16

yw