# simultaneous equations involving e and ln.

• December 13th 2007, 12:01 PM
Oranges&Lemons
simultaneous equations involving e and ln.
urgh, i dont even know where to start. any help would be much appreciated.

i have written the power in brackets because i dont know how to make them superscript, sorry!

solve to 2dp the following simultaneous equations.

e(to the power of 2y) - x + 2 = 0
ln(x+3) - 2y - 1 = 0

thanks :)

ps, the x+3 is not a power, its actually in brackets :P
• December 13th 2007, 12:08 PM
FalconPUNCH!
While I may not be able to help you this moment because I have class in a couple of minutes I will help with the script

$e^{(2y)} - x + 2 = 0$

$ln(x+3) - 2y - 1 = 0$
• December 13th 2007, 12:11 PM
TKHunny
What's the hangup? This one solve directly. Why do you need to specify decimal places?

From the second, y = ½(ln(x+3)-1)

Substituting into the first: $e^{ln(x+3)-1}\;-\;x\;+\;2\;=\;0$

Resolve that exponent and you are almost done!
• December 13th 2007, 12:23 PM
Oranges&Lemons
doesnt that result in cancelling out the x's though?
• December 13th 2007, 12:40 PM
TKHunny
$e^{ln(x+3)-1}\;=\;e^{ln(x+3)}e^{-1}\;=\;\frac{x+3}{e}$

Doesn't look like it. Let the notation help you. No need to guess.