How do you solve #e^x(2e^x-1) = 10#?

1 Answer
Jun 2, 2018

Real solution: #x=ln(5/2)#

Complex solutions:

#x=ln(5/2)+2npii#

#x=ln(2)+(2n+1)pii#

Explanation:

Given:

#e^x(2e^x-1) = 10#

We can treat this as a quadratic in #e^x# and factor it using an AC method:

Transposing and subtracting #10# from both sides, this becomes:

#0 = e^x(2e^x-1)-10#

#color(white)(0) = 2(e^x)^2-(e^x)-10#

#color(white)(0) = (2(e^x)^2-5(e^x))+(4(e^x)-10)#

#color(white)(0) = e^x(2e^x-5)+2(2e^x-5)#

#color(white)(0) = (e^x+2)(2e^x-5)#

So:

#e^x = -2" "# or #" "e^x = 5/2#

If #x# is Real then #e^x > 0#. So the only Real solution to the given equation is given by taking the natural logarithm of both sides of the second equation to find:

#x = ln (5/2)#

Note however that #e^(ipi) = -1#, and hence there are complex solutions:

#x = ln(5/2) + 2npi i#

#x = ln(2) + (2n+1)pi i#

for any integer #n#