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December 22, 2014

December 22, 2014

Posted by **Leanna** on Monday, September 6, 2010 at 11:39am.

- Calculus -
**MathMate**, Monday, September 6, 2010 at 1:06pmsubstitute y=ln(x) ....(1)

to get

5+y=14/y

Solve the resulting quadratic equation in y.

Note that the domain of ln(x) is ℝ+\0, reject the value of y≤0.

Solve for x from (1) above.

- Calculus -
**Leanna**, Monday, September 6, 2010 at 6:45pmok, so i got e^-7 and e^2

does that mean i reject e^-7 because its less than or equal to 0?

- Calculus -
**MathMate**, Monday, September 6, 2010 at 7:40pmYour answers are correct,

x=e^{-7}or x=e².

In fact, the value of y in y=ln(x) can be anything in ℝ, so your answers are correct. The restriction for non-negative values are on x only. My apologies.

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