Question

calculus @solomonzelman

Answer 1

Solve for y: \(\large x=\Large 2^{3-y}\)

Answer 2

\(\color{#000000 }{ \displaystyle x=2^{3-y} }\)

\(\color{#000000 }{ \displaystyle x=2^{3}\times 2^{-y} }\)

\(\color{#000000 }{ \displaystyle x=\frac{8}{2^y} }\)

\(\color{#000000 }{ \displaystyle 2^y=8x^{-1} }\)

\(\color{#000000 }{ \displaystyle \ln(2^y)=\ln(8x^{-1}) }\)

and so on..

Answer 3

that's not calculus

Answer 4

Yes, it isn't:)

Answer 5

Well, I'm taking calculus, and this is my homework. So...

Answer 6

Or, just straight,
\(\color{#000000 }{ \displaystyle x=2^{3-y} }\)

\(\color{#000000 }{ \displaystyle\ln(x)=\ln(2^{3-x}) }\)

Answer 7

I meant, \(\color{#000000 }{ \displaystyle\ln(x)=\ln(2^{3-y}) }\)

Answer 8

The ln part is where I get confused

Answer 9

what is confusing?

Answer 10

You are familiar with logarithmic functions, correct?

Answer 11

\(\large \ln(2^{3-y})=(3-y)\ln 2\) right?

Answer 12

\(\color{#000000 }{ \displaystyle\ln(x)=\ln(2^{3-x}) }\)

\(\color{#000000 }{ \displaystyle\ln(x)=(3-y)\ln(2) }\)

yes, right.

Answer 13

I wrote x again .. sorry

Answer 14

So... y = \(\Large -\frac{\ln(x)}{\ln(2)}-3\) ?

Answer 15

Sorry, +3 I meant

Answer 16

Yes, that is exactly right;)

Answer 17

Awesome, thanks!

Answer 18

\(\color{#000000 }{ \displaystyle y=-\frac{\ln x}{\ln 2}+3 }\)

Answer 19

YW