What is the answer to (1000)/(1+39e^{-0.42x)}=630 ?

The answer to (1000)/(1+39e^{-0.42x)}=630 is x=-(ln(37/2457))/(0.42)

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(1000)/(1+39e^{-0.42x)}=630

\frac{1000}{1 + 39 e ^{- 0.42 x}} = 630

x = - \frac{ln ( \frac{37}{2457} )}{0.42}

Decimal Notation

x = 9.98994 \dots

Solution steps

\frac{1000}{1 + 39 e ^{- 0.42 x}} = 630

Multiply both sides by

1 + 39 e^{- 0.42 x}

\frac{1000}{1 + 39 e ^{- 0.42 x}} (1 + 39 e^{- 0.42 x}) = 630 (1 + 39 e^{- 0.42 x})

Simplify

1000 = 630 (1 + 39 e^{- 0.42 x})

Switch sides

630 (1 + 39 e^{- 0.42 x}) = 1000

Divide both sides by

630

1 + 39 e^{- 0.42 x} = \frac{100}{63}

Move

1

to the right side

39 e^{- 0.42 x} = \frac{37}{63}

Divide both sides by

39

e^{- 0.42 x} = \frac{37}{2457}

Apply exponent rules

- 0.42 x = ln (\frac{37}{2457})

Solve

- 0.42 x = ln (\frac{37}{2457}) : x = - \frac{ln ( \frac{37}{2457} )}{0.42}

x = - \frac{ln ( \frac{37}{2457} )}{0.42}

Verify Solutions:

x = - \frac{ln ( \frac{37}{2457} )}{0.42}

True

The solution is

x = - \frac{ln ( \frac{37}{2457} )}{0.42}

3 = e^{2 x}

5 = a \times b^{- (x - c)} + d

1 6^{3 x - 2} = 3 2^{x + 4}

9^{x} - 2 \times 3^{x + 2} + 81 = 0

8^{x} = 1024

What is the answer to (1000)/(1+39e^{-0.42x)}=630 ?
The answer to (1000)/(1+39e^{-0.42x)}=630 is x=-(ln(37/2457))/(0.42)