How do you solve #log_5(log3x) = 0#?

1 Answer
Alan P.
Dec 15, 2015

Use the equivalence:
#color(white)("XXX")color(red)(log_b(a)=c) hArr color(blue)(b^c=a)#
to determine
#color(white)("XXX")x=10/3#

Explanation:

#log_5(log (3x))=0#

#rArr 5^0 = log(3x)#
#rarrcolor(white)("XXX")log(3x)=1#

#log(3x)=1#
#rArr 10^1=3xcolor(white)("XXXXXXXXXXXXX")#remember the default #log# base is #10#
#rarrcolor(white)("XXX")x=10/3#

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