Question 236518
{{{2^(x+1)=e^x}}} Start with the given equation.



{{{ln(2^(x+1))=ln(e^x)}}} Take the natural log of both sides.



{{{(x+1)ln(2)=x*ln(e)}}} Pull down the exponents using the identity  {{{ln(x^y)=y*ln(x))}}}



{{{(x+1)ln(2)=x*1}}} Evaluate the natural log of 'e' to get 1.



{{{(x+1)ln(2)=x}}} Multiply



{{{x*ln(2)+ln(2)=x}}} Distribute



{{{ln(2)=x-x*ln(2)}}} Subtract {{{x*ln(2)}}} from both sides.



{{{ln(2)=x(1-ln(2))}}} Factor out the GCF 'x'



{{{ln(2)/(1-ln(2))=x}}} Divide both sides by {{{1-ln(2)}}}.



So the solution is {{{x=ln(2)/(1-ln(2))}}}