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Question
solve the equation algebraically.

1) (1.045)^t=2
2) e^x + e^-x = 3

Answer
1.)
b^y = x means log(b)x = y

(1.045)^t = 2
log(1.045)2 = t
t = (log(2))/(log(1.045))
t = 15.7473

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2.)
e^x + e^(-x) = 3
e^x + (1/(e^x)) = 3
((e^x)^2 + 1)/(e^x) = 3
(e^x)^2 + 1 = 3(e^x)
(e^x)^2 - 3(e^x) + 1 = 0

lets think of the (e^x) as x, to make this easier

x^2 - 3x + 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-(-3) ± sqrt((-3)^2 - 4(1)(1))/2(1)
x = (3 ± sqrt(9 - 4))/2
x = (3 ± sqrt(5))/2

or in this case

e^x = (3 ± sqrt(5))/2

log(e)((3 ± sqrt(5))/2) = x

x = ln((3 ± sqrt(5))/2)

x = ±.9624

You can find the answers by going to www.quickmath.com

info found at http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/logs.html

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