Question

Solve exactly.
8^(1-y) = 4^(y+2)

y= ?

Answer 1

Note that \(8 = 2^3\) and \(4=2^2\). So re-writing your equation using this form and taking the \(log_2\) of both sides will probably do nice things.

Answer 2

Or just equating the exponents once you have both sides as a power of two amounts to the same thing.

Answer 3

hmmm not following

Answer 4

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

Answer 5

\[8 = 2^3\]
\[\implies 8^{1-y} = (2^3)^{1-y} = 2^{3(1-y)}\]
\[4 = 2^2\]
\[\implies 4^{y+2} = (2^2)^{y+2} = 2^{2(y+2)}\]
\[\therefore 8^{1-y} = 4^{y+2} \]\[\implies 2^{3(1-y)} = 2^{2(y+2)} \]\[\implies 3(1-y) = 2(y+2)\]

Answer 6

That's right. You're on the track now.

Answer 7

Take the natural log of both sides then take the exponents (1-y) and (y+2) down to the front of each term as factors then gather the y variables and factorise and simplify.

Answer 8

and the 2's cancel out because you are using log 2?

Answer 9

Right

Answer 10

Or just equating exponents.

Answer 11

hmmm I get -1/5

Answer 12

I get postive 1/5

Answer 13

Double check your algebra.

Answer 14

Oh wait. I should check mine. -1/5 is right.

Answer 15

lol

Answer 16

-0.2 or -1/5 is correct. :)

Answer 17

lol was gonna say that what I keep getting