Question

I need help with this logarithm equation on the steps to solve. 400/1+e^-x=350

Answer 1

400 /(1+e^x) = 500?

Answer 2

but not 500 , 350

Answer 3

oh yes - sorry

Answer 4

I got -0.13353192

Answer 5

Multiply through by the denominator to get 400=350+350e^x
Then isolate the variable: 50/350=e^x
Apply the definition of logarithm: x=ln(1/7)

Answer 6

I have the answer, thas not it . @ebnor

Answer 7

\[400/(1+e ^{-x}) = 350\]\[400/1=350e ^{-x}\]\[400/350 = e ^{-x}\]\[\ln 400/350= lne ^{-x}\]\[-0.13353192 =x\] Where did I go wrong?

Answer 8

\[\large \frac{400}{1+e^{-x}}=350\]


\[\large 400=350(1+e^{-x})\]


\[\large 400=350+350e^{-x}\]


\[\large 400-350=350e^{-x}\]


\[\large 50=350e^{-x}\]


\[\large \frac{50}{350}=e^{-x}\]


\[\large \frac{1}{7}=e^{-x}\]


\[\large \frac{1}{7}=\frac{1}{e^{x}}\]


\[\large e^{x}=7\]


\[\large x=\ln(7)\]

Answer 9

The answer is suppose to be 1.95491.

Answer 10

@jim explained it logically , thanks

Answer 11

The exact answer is \[\large x=\ln(7)\] the approximate answer is \[\large x\approx 1.94591014\]

Answer 12

for the ones with negative answers, they lost a negative somewhere

Answer 13

Yeah, I redid it with a few changes and got that too. I didn't carry the "1" over in my first attempt, that's where I went wrong. Can someone help me with my question?

Answer 14

post to the left (if the question is completely different) or ask here (if it's the same or very similar)

Answer 15

woops didnt see the negative x. so i redid it above with a -x

Answer 16

personally, it's better to start a new thread since large threads tend to lag

Answer 17

I've already started one