Question

Hi - I've one equation left but just cannot solution it.
(9^y)/(27^(2y+1))=81 and I know that y=-7/4 but I just can't get the working right - any help/guidance is appreciated - thanks.

Kev.

Answer 1

9=3^2
27=3^3

Answer 2

write everything with the same base
use 3

Answer 3

what fiddle said. get
\[\frac{3^{2y}}{3^{3(2y+1)}}=3^4\]

Answer 4

I was working on that - and got it to 3^2y/3^6y+1 = 3^4 and then knocking the base out to give me 2y/6y+1 = 4 but that seemed to easy

Answer 5

ahh, yep as satellite73 has listed (thanks, I'm on the right track) but now I just don't seem able to balance the thing :(

Answer 6

because it is wrong. you have so subtract the exponents

Answer 7

\[2y-3(2y+1)=4\]

Answer 8

that works out to -7/4

Answer 9

and don't forget the distributive property

Answer 10

\[2y-6y-3=4\] etc

Answer 11

face+palm = doh! Only been on this one for 2hrs and just couldn't get it

Answer 12

thanks folks - feel like such a dork. It's going to teach me to offer to help the kids with their A level Maths homework - I managed the other five they couldn't get - but this one was beyond my teaching of 20yrs ago.

Answer 13

nope - just didn't get the last stage - guess it's tiredness (0030 here in UK) but I'm not seeing how 3^2y/3^6y+1 = 3^4 ends up being 2y-6y-3=4 and that becomes y=-7/4 .... oh well, thanks for the help folks.

Answer 14

(9^y)/(27^(2y+1))=81

\[\frac{(3^2)^y}{(3^3)^{(2y+1)}}=3^4\]
\[\frac{3^{2y}}{3^{3(2y+1)}}=3^4\]

In division (powers of the same base) we subtract the exponents:
\[3^{2y-3(2y+1)}=3^4\]

Compare the exponents of 3 on both sides of the equation above, and get::
\[2y-3(2y+1)=4\]
\[2y-6y-3=4\]
\[-4y=7\]
\[y=-\frac{7}{4}\]