Question 130650
{{{e^(3x+1)=1/(e^(x-2))}}} Start with the given function



{{{e*e^(3x)=1/(e^(x-2))}}} Break up {{{e^(3x+1)}}} to get {{{e^(3x)*e^1=e*e^(3x)}}}



{{{e*e^(3x)=1/(e^x/e^2)}}} Break up {{{e^(x-2)}}} to get {{{e^(x)*e^(-2)=e^x/e^2}}}


{{{e*e^(3x)=e^2/e^x}}} Flip the fraction on the right side



{{{e(e^x)^3=e^2/e^x}}} Rewrite {{{e^(3x)}}} to get {{{(e^x)^3}}}



Now let {{{u=e^x}}}


{{{e*u^3=e^2/u}}} Plug in {{{u=e^x}}} 



{{{e*u^3*u=e^2}}} Multiply both sides by {{{u}}}



{{{e*u^4=e^2}}} Multiply


{{{u^4=e^2/e}}} Divide both sides by {{{e}}}


{{{u^4=e}}} Reduce



{{{(e^x)^4=e}}} Now plug in {{{u=e^x}}}



{{{e^(4x)=e}}} Multiply



{{{e^(4x)=e^1}}} Rewrite {{{e}}} as {{{e^1}}}



Since the bases are equal, the exponents are equal


So {{{4x=1}}}



{{{x=1/4}}} Divide both sides by 4 to isolate x




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Answer: 


So our solution is 



{{{x=1/4}}}