How do you solve #-log x = 4.176#?

1 Answer
Alan P.
Dec 7, 2015

#x~= 6.668xx10^(-5)#

Explanation:

#-log x = 4.176#
#color(white)("XXX") rArr log x = -4.176#

#color(white)("XXX")10^(-4.176) = x#

therefore (using a calculator)
#color(white)("XXX")x = 6.668xx10^(-5)#

Remember
#color(white)("XXX")log_b a = c# means #b^c = a#
and
#color(white)("XXX")#the default base (#b#) for #log# is #10#

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