How do you solve #e^ { x ^ { 2} } = e ^ { 13x } \cdot \frac { 1} { e ^ { 40} }#?

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Answer 1 · 2017-10-26T04:14:56

#x = 8, 5#

#(e^x)^2= e^(13x) * (1/e^40)#

#(e^x)^2 - e^(13x) * (1/e^40)#

#e^(x^2 - 13x) = e^-40#

#x^2 - 13x = -40#

#x^2 -13x +40 = 0#

#(x-8)(x-5) = 0#

#x = 8, 5#