Question

Please helppppppppp I'm drowning in an ocean of confusion!

Answer 1

Not good at math, but will try.

Answer 2

\[(\frac{ 2^{-n} }{ 3 })(\frac{ 3^{-n} }{ 2 })=\frac{ 1 }{ 36 }\] what is the value of n?

Answer 3

I have the answer i just need to know why and how to do the problem.

Answer 4

\(\large\color{black}{ \displaystyle\left( \frac{2^{-n}}{3}\right)\left( \frac{3^{-n}}{2}\right) }\)


\(\large\color{black}{ \displaystyle 2^{-n} \cdot 3^{-1} \cdot 3^{-n} \cdot 2^{-1} }\)


\(\large\color{black}{ \displaystyle 2^{-n} \cdot 2^{-1}\cdot 3^{-1} \cdot 3^{-n} }\)


\(\large\color{black}{ \displaystyle 2^{-n-1} \cdot 3^{-n-1} }\)


\(\large\color{black}{ \displaystyle (2 \cdot 3)^{-n-1} }\)


\(\large\color{black}{ \displaystyle (6)^{-n-1} }\)

Answer 5

So you have:

\(\large\color{black}{ \displaystyle (6)^{-n-1} =6^{-2} }\)

Answer 6

questions?

Answer 7

OH GOD! That was tricky! So basically they converted 6^-2 to 1/36 right?

Answer 8

that might be how they made the problem.
So I am just going backwards,

and you know that for any numbers a, b, c:

if: \(\large\color{black}{ \displaystyle a^b =a^c }\)
then: \(\large\color{black}{ \displaystyle b=c }\)

Answer 9

So in your case,

\(\large\color{black}{ \displaystyle 6^{-n-1} =6^{-2} }\),

then, you can also equate the exponents.

Answer 10

yea i got that part

Answer 11

now, you can solve for n.... can yo?

Answer 12

yea its 1 thanks a lot!!

Answer 13

yes, n=1 is correct. Yw

Answer 14

Hey one last question is there a reason why it doesn't become a positive n when the 2^-n and 3^-n are multiplied?

Answer 15

Like why isn't it a 6^n

Answer 16

you add exponents (if the base is same)

Answer 17

I added exponents, when I had 2^-1 and 2^-n, with 3^-n and 3^-1/

Answer 18

I added exponents, not multiplied.

Answer 19

oh oops sory lmao i was lookin at it wrong

Answer 20

.... :D