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e^x + e^2x = e^3x how to solve this ?=[
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\[e ^{x} + e ^{2x} = e ^{3x}\]
start by substitution w+w^2-w^3=0 w(1+w-w^2)=0
let u=e^x u+u^2-u^3=0 u(1 + u - u^2)=0 u=0 and solve that quadratic using the quadratic formula. don't forget to back substitute and solve for x.
oh !
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\[e^x +e^{2x} =e^{3x}\]\[e^x +e^{2x} -e^{3x}=0\]\[e^x(1 +e^{x} -e^{2x})=0\]\[e^x=0(rejected) \ or\ (1 +e^{x} -e^{2x})=0\]\[ (1 +e^{x} -e^{2x})=0\]\[e^x = \frac{-1 \pm \sqrt{(1)^2 -4(-1)(1)}}{2(-1)}\] .... Can you do it?
that one is better ^^^
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