exponent and log

so far i know that ln(1-y) - ln(y) can be simplified to ln(1-y/y) and the ln can be removed via exponential so now i have :
e^x = 1 - y / y

but am not sure how to form the proved solution. Am confused by how e^-x is formed?

Reply 1

Gregorius

14

Original post by wagon23

so far i know that ln(1-y) - ln(y) can be simplified to ln(1-y/y) and the ln can be removed via exponential so now i have :
e^x = 1 - y / y

but am not sure how to form the proved solution. Am confused by how e^-x is formed?

Try solving the slightly simpler equation

z = \frac{1 - y}{y}

for y. Then substitute z = e^x and notice that 1/e^x = e^(-x)

Reply 2

wagon23

OP

9

Original post by Gregorius

Try solving the slightly simpler equation

z = \frac{1 - y}{y}

for y. Then substitute z = e^x and notice that 1/e^x = e^(-x)

so after re-arranging and making y the subject I get y = 1/ z + 1? subbing e^x now you get e^-x + 1 = y? how is the numerator formed?

Reply 3

goldenusername

15

Is this from mei?

After you rearrange to get 'y' as your subject, you should end up with y=1/e^x+1. And then you multiply your fraction by e^-x/e^-x.

Reply 4

Gregorius

14

Original post by wagon23

so after re-arranging and making y the subject I get y = 1/ z + 1? subbing e^x now you get e^-x + 1 = y? how is the numerator formed?

You should get

y = \frac{1}{1+z}

Reply 5

wagon23

OP

9

Original post by Gregorius

You should get

y = \frac{1}{1+z}

oh i see where i went wrong thanks

Reply 6

wagon23

OP

9

Original post by goldenusername

Is this from mei?

After you rearrange to get 'y' as your subject, you should end up with y=1/e^x+1. And then you multiply your fraction by e^-x/e^-x.

cie. and thanks for the help

exponent and log

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