Question

How to get from
5(2^(n-1) + 4 * 3^(n-1)) - 6(2^(n-2) + 4 * 3^(n-2))
to
(5 * 2 - 6)2^(n-2) + (5 * 4 * 3 - 6 * 4)3^(n-2)
by simplifying the initial expression?

Answer 1

what have you worked out so far?

Answer 2

Nothing this isn't the problem it's part of a solution and I'm years out of practice in math and I just have absolutely no idea how to get from that one piece to the next.

Answer 3

factoring is key ... but so is some basic exponent rules

Answer 4

distribute, collect like terms 2^this and 3^that

and factor out what needs to be factored to turn it

Answer 5

yeah i know that but i don't know HOW. So what, the first half of the first expression would be 10^(n-1) + 20 * 15^(n-1)? Is that correct? Then deal with the negative exponents or what

Answer 6

as long as your willing to participate, im willing to help out ... your first reply was sufficient

Answer 7

lets keep the actual multiplication in parts for the moment, jsut show the distribution itself

5(2^(n-1)) + 5(4)(3^(n-1)) - 6(2^(n-2)) + 6(4)(3^(n-2)))

now we move things about since our end reult wants to be in terms of 2^ and 3^

Answer 8

gathering like terms

5(2^(n-1))- 6(2^(n-2)) + 6(4)(3^(n-2)))+ 5(4)(3^(n-1))

now we are in position to factor out what needs to be facotred right?

Answer 9

ok yeah

Answer 10

no wouldn't it be - 6(2^(n-2)) - 6(4)(3^(n-2))) from your prev post

Answer 11

since it's -6 * (4)(3^(n-2)))

Answer 12

we know we want to factor out 2^(n-2) and 3^(n-2)

[5(2^(n-1))/(2^(n-2)) - 6] 2^(n-2)
+ [6(4)+ 5(4)(3^(n-1))/3^(n-2) ] 3^(n-2)

Answer 13

the like terms we are looking for is not 6, but looking at what we want ... its going to have to be the 3^ parts

Answer 14

talking about the distribution 5(2^(n-1)) + 5(4)(3^(n-1)) - 6(2^(n-2)) + 6(4)(3^(n-2))) should be 5(2^(n-1)) + 5(4)(3^(n-1)) - 6(2^(n-2)) - 6(4)(3^(n-2))) instead or am i wrong

Answer 15

lets look at this in the coding .....
\[5~[2^{n-1} + 4~3^{n-1}] - 6~[2^{n-2} + 4~3^{n-2}]\]
distribute, and yes, it looks like i dropped that negative, my err
\[5~2^{n-1} + 5~4~3^{n-1} - 6~2^{n-2} - 6~ 4~3^{n-2}\]
reaarange the like terms
\[5~2^{n-1}- 6~2^{n-2} + 5~4~3^{n-1} - 6~ 4~3^{n-2}\]
factor out the required parts
\[[5~\frac{2^{n-1}}{2^{n-2}}- 6]~2^{n-2} + [5~4~\frac{3^{n-1}}{3^{n-2}} - 6~ 4]~3^{n-2}\]

Answer 16

exponent rule:

b^n/b^m = b^(n-m) so that should get it to what you were looking for

Answer 17

feels like the site is starting to breakdown again. notifs and posts are getting slowed down

Answer 18

sok slow seems to be a better speed for my brain

Answer 19

because how did you get from rearranging the like terms to "factor out the required parts" what happened there

Answer 20

with the exponential terms parts

Answer 21

the rearrange is just to get all the parts nest to each other that have the required factor that will be pulled out

commutative property at work is all.

the end results shows us what to work towards: () 2^ and () 3^

so i simply arranged them and pulled out the needed factor

are you asking why a divided as i did in the last step?

Answer 22

yes this whole thing
factor out the required parts
[5 2n−12n−2−6] 2n−2+[5 4 3n−13n−2−6 4] 3n−2

Answer 23

well i just copy pasted it didn't paste right but the last part of your post

Answer 24

lets make sure we are in step for the rearranging ....
\[5~2^{n-1}- 6~2^{n-2} + 5~4~3^{n-1} - 6~ 4~3^{n-2}\]
its good for you up to here right?

Answer 25

refresh gets rid of the odd marks but then i was stuck in a death spiral refresh page load lol

Answer 26

the concept i want to use here is that 1, times anything, changes nothing.

1(8) = 8
1(2) = 2
1(15) = 15

so, im thinking we need to multiply by a useful 'form' of 1

anything divided by itself (except 0/0) is equal to 1 sooo

\[5~2^{n-1}- 6~2^{n-2} + 5~4~3^{n-1} - 6~ 4~3^{n-2}\]
\[5~2^{n-1}~1- 6~2^{n-2} + 5~4~3^{n-1}~1 - 6~ 4~3^{n-2}\]
\[5~2^{n-1}~\frac{2^{n-2}}{2^{n-2}}- 6~2^{n-2} + 5~4~3^{n-1}~\frac{3^{n-2}}{3^{n-2}} - 6~ 4~3^{n-2}\]
and factor out the required parts
\[[5~\frac{2^{n-1}}{2^{n-2}}- 6]~2^{n-2}~+~[5~4~\frac{3^{n-1}}{3^{n-2}} - 6~ 4]~3^{n-2}\]

Answer 27

the last step is just simplifying the fractions .... using the exponent rule
\[[5~\frac{2^{n-1}}{2^{n-2}}- 6]~2^{n-2}~+~[5~4~\frac{3^{n-1}}{3^{n-2}} - 6~ 4]~3^{n-2}\]
\[[5~2^{n-1-n+2}- 6]~2^{n-2}~+~[5~4~3^{n-1-n+2} - 6~ 4]~3^{n-2}\]
\[[5~2^{1}- 6]~2^{n-2}~+~[5~4~3^{1} - 6~ 4]~3^{n-2}\]
\[[5~2- 6]~2^{n-2}~+~[5~4~3 - 6~ 4]~3^{n-2}\]