Asked by Abraham
                solve the equations for x
a) e^2x - e^x - 30 = 0
b)1-3/n^x = 5
            
        a) e^2x - e^x - 30 = 0
b)1-3/n^x = 5
Answers
                    Answered by
            Damon
            
    let z = e^x
then
z^2 - z - 30 = 0
(z-6)(z+5) = 0
so
z = 6 or -5
if z = e^x = 6
x = ln 6 look it up
if z = -5
x = ln -5 impossible so only ln 6 is solution
    
then
z^2 - z - 30 = 0
(z-6)(z+5) = 0
so
z = 6 or -5
if z = e^x = 6
x = ln 6 look it up
if z = -5
x = ln -5 impossible so only ln 6 is solution
                    Answered by
            Damon
            
    Are you sure this is not a typo?
1-3/n^x = 5
-3/n^x = 4
n^x = -3/4
ln n^x = -ln .75
x ln n = -ln.75
x = -ln.75/ln n
x = .2877/ln n
    
1-3/n^x = 5
-3/n^x = 4
n^x = -3/4
ln n^x = -ln .75
x ln n = -ln.75
x = -ln.75/ln n
x = .2877/ln n
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