Question

how to solve this indice simultaneous eqn?

3^x-1 = 9x3^y ------- (1)
2^x + 2^y = 9 ------- (2)

Answer 1

is that (9x^3)^y?

Answer 2

its 9 times 3^y.

Answer 3

oh okk

Answer 4

thanks :)

Answer 5

and is that 3^x -1 or 3^(x-1)

Answer 6

3 to the power of (x-1).

Answer 7

change it so it all has the same base 3^(x-1) = 3^3y

Answer 8

ok then the first equation becomes

3^(x-1) = 3^(2+y)

Answer 9

ok.i got the first one.but the second equation has different bases of 2 and 9.

Answer 10

actually i am stuck in the second equation about how to change it to same base.

Answer 11

\[2^{x} + 2^{y} = 9\]

Answer 12

x-1=2+y
x=3+y

nw substitute

2^(2+y) + 2^y =9
4*2^y +2^y =9
2^y(1+4)=9
2^y=9/5

Answer 13

this is where i got stuck.

Answer 14

y=0. x=3

Answer 15

thats the answer written in the book.i just need to know how to solve it.

Answer 16

right. x=3+y. put this in the second equation. so 2^(3+y)+2^y=9. this is 9.2^y=9. i.e 2^y=1. so y=0.
this gives x=3, sonce x=3+y

Answer 17

*since

Answer 18

this is 9.2^y=9. i.e 2^y=1. so y=0.. can you explain this part bro?

Answer 19

i understand the substitution part.but not what happens afterwards.

Answer 20

2^3=8. so 2^(3+y) + 2^y =9 simplifies to 8.2^y + 2^y =9. this reduces to 9. 2^y =9. get it?

Answer 21

bro i am really sorry but u mind explaining how you get it again?really appreciate it man.

Answer 22

2^3 = 8 but how come it becomes 8.2?

Answer 23

fine, ill walk you through it. 3\[3^{x-1}=3^{2+y}\]
right? equate the exponents.. thsi gives \[x=3+y\]
put this in the second equation \[2^{3+y} + 2^ y =9\]

this is \[8\times 2^y + 2^y = 9\]

which is \[9 \times @^y = 9\]

y=0. x=3

Answer 24

OH I GOT IT BRO.I just got confused by where 8.2 came from.

Answer 25

Really a great help man..Wish you were my math teacher. :)

Answer 26

Thanks again bro :)

Answer 27

see ya dude :)