Question

logx(1/49) = -2

Answer 1

Take the\[\sqrt[-2]{1/49}\] negative square root of (1/49)

Answer 2

Okay... I put that in the calculator?

Answer 3

yeah that's how you solve the base log of something given those numbers

Answer 4

I got -.14

Answer 5

you have to use the function on your calculator that looks like\[\sqrt[n]{x}\]

Answer 6

\[x^ {-2}=\frac{ 1 }{ 49 }\]

Answer 7

That works too, then you would just take both sides and raise them to the exponent of -1, giving you
\[x ^{2} = 49\]

Answer 8

the -2 isn't exponent?

Answer 9

What you were doing wrong I think when using my method is you had to put n = (-2) in brackets when using \[\sqrt[n]{x}\]

Answer 10

I put - square root of (1/49), I got -.14 , I just made it short by 2 decimals.

Answer 11

So in my calculator, I would put (-2)sqrt(1/49). Use brackets. You should get 7.

Answer 12

I can't remember how to put the -2 in the sqrt

Answer 13

Don't use the normal sqrt function, there should be separate one that maybe looks like\[\sqrt[x]{?}\]

Answer 14

the only square root that I see is above the x^2 button.

Answer 15

Hm, I guess your calculator does not support that function. That's fine. Just think of it as
\[\sqrt{(1/49)^{-1}}\] That should make sense

Answer 16

I just figured it out I was suppose to press math button then it'll show up the list.

Answer 17

ah okay, do you get 7 now?

Answer 18

with the \[\sqrt{(1/49)^{-1}}\] method , I did

Answer 19

Yeah that's fine, it means the same thing. I didn't realise you couldn't find the other square root function

Answer 20

Yeah, this is the graphing calculator, and I didn't remember how to do that cause I did that last semester, but doing the review now before the final exam

Answer 21

Oh okay, so you do you understand it now?

Answer 22

What about log6 x = 2?