Question

Two in one :) Solve the equation.

In (x+5)= In (x-1) - In (x+1)

&

400/1+e^-t = 350

Answer 1

\[\frac{x-1}{x+1}=x+5\]\[(x+1)(x+5)=x-1\]\[x^2+5x+x=x-1\]

Answer 2

\[\ln(x+5)=\ln(\frac{x-1}{x+1})\]

\[x+5=\frac{x-1}{x+1}\]

Answer 3

\[x^2+5x+6=0\]

Answer 4

for the second problem I would start by multiplying the whole equation by 350e^t since e^-t is really 1/e^t

Answer 5

(x+2)(x+3)=0

Answer 6

wait

Answer 7

\[140,000e ^{t}+350=e ^{t}\]

Answer 8

I do right

Answer 9

wait a sec the second problem has no real solution to it

Answer 10

err t=

Answer 11

there is a chance he meant to write it as
\[\frac{400}{1+e^{-t}}=350\]

Answer 12

oh right that makes sense

Answer 13

400=350(1+e^{-t})
400=350+350e^{-t}
400e^t=350e^t+350
400e^t-350e^t=350
50e^t=350
e^t=7
and so on...

Answer 14

t = ln(7)

Answer 15

\[\frac{x-1}{x+1}=x+5\](x+1)(x+5)=x-1\[x^2+5x+6=0\]\[(x+2)(x+3)=0\]
x=-2
x=-3

Answer 16

i keep trying to solve problems that are written wrong

Answer 17

in all fairness it wouldnt be logical for 400/1 to appear in a problem

Answer 18

You guys are awesome !! thanks

Answer 19

\[ \log_{b}\sqrt[3]{x ^{2\sqrt{y}}}\div z ^{3} \]

anyone?? The z^3 is divided under the radical

Answer 20

this is not a problem. this is an expression

Answer 21

btw what a weird way to solve
\[\frac{400}{1+e^{-t}}=350\]

Answer 22

ok maybe not