Question 109797
{{{2^x+1=1/32}}}


Please read my lesson---An overview to the laws of logarithms


{{{2^x+1-1=1/32-1}}}
{{{2^x+0=-31/32}}}
{{{2^x=-31/32}}}
{{{log(2^x)=log(-31/32)}}}
{{{xlog(2)=log(-31)-log(32)}}}


WAIT!!

{{{log(-31)}}} is IMPOSSIBLE>


Thus, and therefore, there is no solution.:)

Power up,
HyperBrain!