Question 936334
{{{7^(2x+1)=4^(x-2)}}} .....since {{{7^(2x+1)=7^(2x)*7}}} and {{{4^(x-2)=4^x/4^2}}}, we have


{{{7^(2x)*7=4^x/4^2}}}


{{{7^(2x)*7=4^x/4^2}}} 


{{{7^(2x)/4^x=1/(7*4^2)}}}  .........use a logarithm


{{{log((7^(2x)/4^x))=log((1/(7*4^2)))}}}


{{{log((7^(2x)))-log((4^x))=log(1)-log((7*4^2))}}}


{{{2x*log((7))-x*log((4))=log((1))-log((7))+2log((4))}}} ...{{{log((1))=0}}}


{{{x(2log((7))-log((2^2)))=2log((2^2))-log((7))}}}


{{{x(2log((7))-2log((2)))=4log((2))-log((7))}}}


{{{x=(4log((2))-log((7)))/(2(log((7))-log((2))))}}}