Question

calculate the value of y for which 2log(3)y-log(3)(y-4)=2
(the 3 is the base in both cases i just didnt know how to show subscript on my computer)

Answer 1

\[2 \log_3(y)-\log_3(y-4)=2\]

Answer 2

use power rule first for the 2log(y) thing

Answer 3

then you will want to use quotient rule for the the part that results from that

Answer 4

useful rules here:
\[r \log_a(y)=\log_a(y^r) \text{ power rule } \\ \log_a(x)-\log_a(y)=\log_a(\frac{x}{y}) \text{ quotient rule } \\ \log_a(w)=q \text{ implies } a^q=w \]

Answer 5

ok so i have log(3)(y^2/y+4)=2

Answer 6

whats next?

Answer 7

the last thing I wrote is next

Answer 8

\[\log_a(w)=q \text{ implies } a^q=w\]

Answer 9

and i think you meant
\[\log_3(\frac{y^2}{y-4})=2\]

Answer 10

but anyways notice that the input becomes the output
and the output becomes the input
when using the equivalent form I just mentioned

Answer 11

that is
\[\log_3(\frac{y^2}{y-4})=2 \text{ implies } 3^{2}=\frac{y^2}{y-4} \\ \text{ just used } \log_a(w)=q \text{ implies } a^q=w\]

Answer 12

oh great thanks i got it to a quadratic of y^2-9y-36=0 (i meant to say (y+4) not (y-4) at the begining

Answer 13

oh ok

Answer 14

solution of y=12

Answer 15

\[3^2=\frac{y^2}{y-4} \\ 9=\frac{y^2}{y-4} \\ 9(y-4)=y^2 \\ 9y-36=y^2 \\ 0=y^2-9y+36\]

Answer 16

oops i forgot to replace y-4 with y+4 lol sorry

Answer 17

i was trying to check your quadratic

Answer 18

\[9(y+4)=y^2 \\ 9y+36=y^2 \\ 0=y^2-9y-36\]
ok your quadratic looks good

Answer 19

and I guess you factored it maybe
(y-12)(y+3)=0

Answer 20

and you knew y=12 would only work
because y=-3 would make the insides of the log negative
and we can't have that

Answer 21

good job @jamie123