Question

satellite73, I need help

Answer 1

Satellite is enjoying a cold beer.

Answer 2

thats a huge problem there

Answer 3

any one?

Answer 4

\[3^y = \sqrt{2^{y-1}}\]

Answer 5

what are we looking to do with it?

Answer 6

Rewrite RHS\[(2^{y-1})^{1/2}\]

Answer 7

sole for y

Answer 8

Then take ln of both sides.

Answer 9

ylog(3)= (y-1)/2log(2)
I know from here,I can't do any farther

Answer 10

\[y \log(3)=(\frac{y-1}{2})\log(2)\]
nex?

Answer 11

Move them over to one side, I usually use ln, I don't know if it makes a difference. Then ln subtraction becomes division.

Answer 12

\[2ylog(3)=(y-1)\log(2)\]

Answer 13

from here how I do next?

Answer 14

No. Subtract RHS, from both sides.

Answer 15

\[3^{2y}=2^{y-1}\]
\[2y\ln(3)=(y-1)\ln(2)\]
\[2y\ln(3)=y\ln(2)-\ln(2)\]
\[2y\ln(3)-y\ln(2)=-\ln(2)\]
\[y(2\ln(3)-\ln(2))=-\ln(2)\]

Answer 16

how am i doing so far?

Answer 17

can u do more?

Answer 18

I think you left out the 1/2 or the radical

Answer 19

one more step yes? divide both sides by
\[2\ln(3)-\ln(2)\]

Answer 20

wait did i make a mistake? hold on i may have. need another beer anyway

Answer 21

Oh, you transferred it to next side. No problem.

Answer 22

oh no. i squared both sides as step one. whew. that is why i got
\[3^{2y}=2^{y-1}\]

Answer 23

but if you are not scared of fractions then use 1/2. why not?

Answer 24

oh i see you have it all worked out. sorry to butt in. it is algebra from
\[2y \ln(3)=(y-1)\ln(2)\] multiply out, put all the y's on one side, factor and divide. you had it i should keep quiet

Answer 25

next?

Answer 26

I don't know from here

Answer 27

last step is to divide. you have
\[y(2\ln(3)-\ln(2))=-\ln(2)\] everything is a number except y. divide to get
\[y=\frac{-ln(2)}{2\ln(3)-\ln(2)}\]

Answer 28

you can write this in many different ways using the property of logs, but i wouldn't bother. for example
\[2\ln(3)=\ln(9)\]
\[\ln(9)-\ln(2)=\ln(\frac{9}{2})\]and so on