Question

solve for x in terms of y

y=2^x-2^-x/2^x+2^-x

Answer 1

help please

Answer 2

\[
y=\frac{2^x-2^{-x}}{2^x+2^{-x}}
\]

Answer 3

yup

Answer 4

Do you know about \(\tanh(x)\)?

Answer 5

\[
\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}
\]

Answer 6

no it is a part of logarithm

Answer 7

\[
\tanh(x\ln2) = \frac{e^{x\ln 2}-e^{-x\ln 2}}{e^{x\ln 2}+e^{-x\ln2}} =\frac{2^x-2^{-x}}{2^x+2^{-x}}
\]

Answer 8

You want to do this with logarithm?

Answer 9

yes not tangent

Answer 10

Well if you have \[
\log_2(y)=\log_2(2^x-2^{-x})-\log_2(2^x+2^{-x})
\]How would you proceed?

Answer 11

Hyperbolic tangent is the only way I know to do this.

Answer 12

Well, you can multiply numerator and denominator by 2^x:
\[y=\frac{2^x(2^x-2^{-x})}{2^x(2^x+2^{-x})}=\frac{2^{2x}-1}{2^{2x}+1}\]
then let u=2^(2x) so that \(y=\frac{u-1}{u+1}\).
You then solve for u in terms of y, then put x back:
\[y(u+1)=u-1\\
uy+y=u-1\\
uy-u=-1-y\\
u(y-1)=-(y+1)\\
u=-\frac{y+1}{y-1}\\
2^{2x}=-\frac{y+1}{y-1}\]
I hope you can take it from here. :)

Answer 13

cross multiplication

Answer 14

No. Your goal was to write y in terms of x, right? And right now, x is already nearly alone. You just have to do something so that the x is not an exponent anymore.
What is the inverse of exponentiation?

Answer 15

i dont know exponentiation

Answer 16

Isn't your topic about logarithms?

Answer 17

wait

Answer 18

the inverse of exponentiation is square root

Answer 19

No, I mean, say you have \(2^x=y\), it is true that \(x=\log_2(y)\), right?

Answer 20

yeah right

Answer 21

So it's true that\[2x=\log_2{\left(-\frac{y+1}{y-1}\right)}\]right?

Answer 22

it strue thats the final answer ?

Answer 23

Not yet. You have to divide both sides by 2 to finally get the final answer.

Answer 24

how?@blockcolder

Answer 25

@blockcolder

Answer 26

Just divide:
\[\frac{2x}{2}=\frac{1}{2}\log_2{\left(-\frac{y+1}{y-1}\right)}\]
So that you get
\[x=\frac{1}{2}\log_2{\left(-\frac{y+1}{y-1}\right)}\]