Question 949837
I'm going to assume the equation is {{{5^(2x+1)=33}}}



{{{5^(2x+1)=33}}}



{{{log((5^(2x+1)))=log((33))}}} Apply logs to both sides



{{{(2x+1)log((5))=log((33))}}}



{{{2x*log((5))+log((5))=log((33))}}}



{{{2x*log((5))=log((33))-log((5))}}}



{{{x*2*log((5))=log((33))-log((5))}}}



{{{x=(log((33))-log((5)))/(2*log((5)))}}}



{{{x=0.58625114844549}}} Use a calculator here (this is approximate)



{{{x=0.5863}}} Round to 4 decimal places (this is approximate)