stuck with this one
solve for x: 16^x-5*8^x=0
Or for log to any base, x log(16)= log(5)+ x log(8) so x(log(16)- log(8))= x(log(16/8))= x(log(2))= log(5) and x= log(5)/log(2).
Another way to do this is, since [tex]8^x[/itex] is never 0, to divide both sides by $\displaystyle 8^x$:
$\displaystyle \frac{16^x}{8^x}= \left(\frac{16}{8}\right)^x= 2^x= 5$
and then $\displaystyle log(2^x)= xlog(2)= log(5)$ so $\displaystyle x= \frac{log(5)}{log(2)}$