Two thinks to remember here:
[1]#color(white)("XXX")color(red)(log_b p - log_b q = log_b p/q)#
[2]#color(white)("XXX")color(blue)(log_b s = t hArr b^t = s)#
Therefore
#color(white)("XXX")color(red)(log_(10) (8x)-log_(10)(1+sqrt(x)) =2)#
means
#color(white)("XXX")color(red)(log_(10) ((8x)/(1+sqrt(x))) =2)#
and
#color(white)("XXX")color(blue)(log_(10) ((8x)/(1+sqrt(x)))=2)#
means
#color(white)("XXX")color(blue)(10^2=(8x)/(1+sqrt(x))#
From this we can get:
#color(white)("XXX")8x=100(1+sqrt(x))#
#color(white)("XXX")2x=25+25sqrt(x)#
#color(white)("XXX")2x-25sqrt(x)-25=0#
If we (temporarily) let #p=sqrt(x)#
#color(white)("XXX")2p^2-25p-25=0#
then applying the quadratic formula
#color(white)("XXX")p=(25+-sqrt((-25)^2-4 * 2 * (-25)))/(2 * 2)#
and with some simplification
#color(white)("XXX")p=(25+-5sqrt(33))/4#
(note, however the negative version can be eliminated as extraneous since #p=sqrt(x) rarr p >=0# for any #x in RR#)
Therefore
#color(white)("XXX")sqrt(x)=(25+5sqrt(33))/4#
and
#color(white)("XXX")x=((25+5sqrt(33))/4)^2 ~~180.3838#
(using a calculator for this last step)
