Question

@agent0smith so if that's the a, does that mean that the answer is Logx^3(9y^2)/z?

Answer 1

Yep.

\[\Large \log(x^3)(9y^2)-\log z = \log \frac{ 9x^3y^2}{ z }\] (that's the same answer, just rearranged a bit)

Answer 2

ahhhhhhhhhhhhhhhhh!!!!!!! i can't believe it!! yess!!

Answer 3

is the y intercept for log(3,2x+1) (0,0)??

Answer 4

i didn't know how to write that, it's log, base 3, (2x+1)

Answer 5

i made it equal to y.

Answer 6

y=log, base 3, (2x+1)

Answer 7

\[\Large y = \log_3 (2x+1)\]

To find the y intercept, plug in x=0

Answer 8

a website said to bring the base over and make y an exponet so then it was 3^y=2x+1

Answer 9

then I made y = 0, and any number to the 0 power is 1

Answer 10

so then I had 1 =2x+1 and ended up with 0=2x, divide both sides and i have 0.

Answer 11

i have never done that before so i'm hoping it's right.

Answer 12

Yes, it would be (0,0). But it's faster if you plug in x=0

\[\Large y = \log_3 (2*0+1) = \log_3 (1) = 0\]

Answer 13

yyaayyy!!

Answer 14

can you explain that a little? how did you get 0 that way?

Answer 15

Log of 1 is equal to zero, no matter the base.

Answer 16

ohhhhh...cool.

Answer 17

what is Find f^-1(x) for f(x)=6x-5? do i just make it y=(6x-5)/-1

Answer 18

That means you have to find the inverse function.

y = 6x-5
swap the x's and y's...
x = 6y - 5

now rearrange it till you get y=... That'll be the inverse, the f^-1(x)

Answer 19

ohhhhhh....

Answer 20

so simple