logarithmic

[sarey126]

Problem:

in solving the problem, [(e^x^+e^-x^)/(2)]=3 ,

I arrived at u^2^-6u+1=0 (using "u" substitution of course), but cannot get any further. I am also just assuming the work I have done sofar is correct. Thank you!

*assume that numbers and quantities between ^...^ are a superscript and those between _...._ are subscripts*

 

[stapel]

The trick for this one (and it is a "trick") is to multiply through by e^x to get a quadratic-type equation. Here's what it looks like:

. . . . .[e^x + e^-x] / 2 = 3

. . . . .e^x + e^-x = 6

Now multiply through by e^x:

. . . . .[e^x]² + 1 = 6e^x

. . . . .[e^x]² - 6e^x + 1 = 0

You'd have no trouble, I'm sure, with:

. . . . .y² - 6y + 1 = 0

Plug-n-chug with the Quadratic Formula, right? So do that here, but you'll be solving for what e^x equals, instead of what y equals. Then use logs to finish the solution.

Eliz.