How do you this problem algebraically?

Question

e^x + e^-x = 10

I solved it graphically, but I have to know how to do it algebraically.

I have no idea how to attack this, I figured I'd take the natural log of all three to get an answer:

lne^x + lne^-x = 10
x - x = ln10
0 = ln10

That's obviously not right. How do I approach this? Thanks.

Answer 1

let z=e^x
z+z^-1=10
z^2+1=10z (Don't you love it when a quadratic suddenly appears?!)
z^2-10z+1=0
using the formulea x=-b+/-root(r2-4ac)/2

z=[10+ root(96)]/2

z=[10- root(96)]/2
replace z with e^x

x=ln [10 (+/-) root(96)]/2

x=2.29243167
or
x=-2.29243167


Please check my working it's exceedingly difficult to do algebra off a computer screen wher eyou can't do power easily etc.

Answer 2

When you took the natural log, you forgot to take the natural log of BOTH sides. You didn't take the log of ten. Furthermore, the natural log is not a linear operator which means that

ln(A+B) is not the same as ln(A)+ln(B). So the right hand side is NOT ln(e^x)+ln(e^(-x)).

So now, taking the log is not the best way to do this problem.

Just do what the person above me did.

Answer 3

e^(2x) + 1 = 10 e^x
e^(2x) - 10 e^x + 1 = 0
Let y = e^x
y ² - 10 y + 1 = 0
y = [10 ± √96 ] / 2
y = [ 10 ± 4√6 ] / 2
y = 5 ± 2√6
e^x = 5 ± 2√6
x = ln [5 ± 2√6 ]

Answer 4

I don't even know that and I'm in algebra at my school! Try searching the internet on algebra math problems.