How do you solve #e^ { 7x } - 8e ^ { 5x } - 9e ^ { 3x } = 0#?

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Answer 1 · 2017-02-13T08:50:31

#x=ln3#

1) take out any common factors

#e^(7x)-8e^(5x)-9e^(3x)=0#

#=>e^(3x)(e^(4x)-8e^(2x)-9)=0#

2) #" the bracket is a quadratic in " e^(2x)" so factorise it as such"#

#=>e^(3x)(e^(4x)-8e^(2x)-9)=0=>e^(3x)(e^(2x)-9)(e^(2x)+1)=0#

3) now solve.

#e^(3x)(e^(2x)-9)(e^(2x)+1)=0#

#e^(3x)=0=>" no real roots"#

#e^(2x)+1=0=>e^(2x)=-1, =>"no real roots"#

#e^(2x)-9=0=>e^(2x)=9, => "root(s) exist(s)"#

#e^(2x)=9=>2x=ln9#

#x=1/2ln9=x=ln9^(1/2)#

#:.x=ln3#