How do solve #4logx = 24.8#?

1 Answer
Aug 29, 2015

Assuming your #log# is base 10:
#4log(x) = 24.8 color(white)("XXX")rArrcolor(white)"XXX"x=1584893#

Explanation:

#4log(x) = 24.8#
#rarrcolor(white)("XXX")log(x)= 24.8/4 = 6.2#

In general
#color(white)("XXX")log_b(x)=ycolor(white)("XXX")"means"color(white)("XXX")b^y=x#

So
#color(white)("XXX")log_10(x)=6.2color(white)("XXX")"means"color(white)("XXX")10^(6.2) =x#
and (with the aid of a calculator)
#color(white)("XXX")x=1584893#