Question

9^(x+1)=27^(x-1)

Can someone PLZ explain how to solve this?

Answer 1

write 9 and 27 as powers of 3

Answer 2

then use the rule\[(x^a)^b=x^{ab}\]

Answer 3

9=3^?
27=3^?

Answer 4

please talk to me so I can help you...

Answer 5

i'm still not understanding. What do u mean write 9 & 27 as powers of 3?

Answer 6

3 to what power equals 9 ?

Answer 7

3^2=9

Answer 8

yes, good :)
and 27=3^?

Answer 9

3^3

Answer 10

good, so we have\[\large 9^{x+1}=27^{x-1}\]\[\large(3^2)^{x+1}=(3^3)^{x+1}\]and now we use the rule\[\large(x^a)^b=x^{ab}\]

Answer 11

ok now how do i use that rule?

Answer 12

in your case we apply the rule on the left as\[x=3,~a=2,~b=x+1\]so we get\[\large(x^a)^b=x^{ab}\]\[\large(3^2)^{x+1}=3^{2(x+1)}=3^{2x+2}\]try to meditate on that until it makes sense...

Answer 13

on the other side we do the same thing with\[\large(3^3)^{x-1}=?\]do this one yourself based on the way I did the other side above

Answer 14

I got \[3^{3x+4}\] is that right?

Answer 15

not quite\[\large(3^3)^{x-1}=3^{3(x-1)}=?\]

Answer 16

\[3^{3x-3}\]?

Answer 17

yes, very good :)

Answer 18

so now look at what we have on both sides...

Answer 19

U should make the bases same in order to make the exponents same and then solve for x...... write 27 as 3^3(x-1) and 9 as 3^2(x+1) now u can see..... 3 is the base on both the sides and if the bases are the same, then the exponents must be equal..... so...... 2(x+1)=3(x-1).... solve for x now.....

Answer 20

\[\large 9^{x+1}=27^{x-1}\]\[\large(3^2)^{x+1}=(3^3)^{x-1}\]\[3^{2x+2}=3^{3x-3}\]now you can take the logarithm of both sides, tight?

Answer 21

\[3^{2x+2}=3^{3x-3}\] Then i did 2x+2=3x-3 & got x=-6.
Is that right?

Answer 22

I was just about to say...there's no need to take logs of both sides.

Answer 23

2x+2=3x-3
^^correct, but you did not solve this right

Answer 24

@Hero I suppose you mean to say that you can just say "the exponents have to be equal" and look at it that way?

Answer 25

yeah @TuringTest, I explained that to her earlier that's why she knew to set the exponents equal, but her value of x is incorrect.

Answer 26

right, gotchya

Answer 27

one sec et me try again

Answer 28

so @joyce153 try again to solve
2x+2=3x-3

Answer 29

I never said that her value of x was correct. I simply said there was no need to take logs

Answer 30

I know, I got it, thanks. I only made the point about taking logs because I assumed she was in a section in which one is learning to use logarithms

Answer 31

is x=-5?

Answer 32

closer but still a bit off
careful with your signs

Answer 33

How did you get the negative?

Answer 34

You should probably show all your steps

Answer 35

ok i did 2x+2=3x-3
then i did 2x-3x+2=3x-3x-3=-x+2=-3
-x+2-2=-3-2=-x=-5

Answer 36

oh nevermind

Answer 37

it just confuses when you write it that way
but you have everything right anyway
-x=-5
so x=?

Answer 38

You want to simplify in such a way where the value of x ends up being positive. If it is negative, then you're not done solving for x.

Answer 39

a neg & a neg = a pos. right? so it be x=5?

Answer 40

yes

Answer 41

Here's how I usually simplify these @joyce153,

2x+2=3x-3


I want x term to be positve so I subtract the smaller term, 2x from both sides, then add 3 to both sides to get

3 + 2 = 3x - 2x
5 = x

That avoids having to deal with negatives at the end.

Answer 42

:( i can only give one of u a meddle but THX @Hero & @TuringTest !!

Answer 43

indeed, and you should write your work thee way Hero has shown above
much easier to follow for both you and the person grading your work

Answer 44

ok!