How do you solve #2log_5x=log_5 9#?

1 Answer
Alan P.
Oct 15, 2016

#x=3#

Explanation:

Remember the general relation:
#color(white)("XXX")log_b (a^c)=c * log_b (a)#

Therefore
#color(white)("XXX")2log_5(x)=log_5(x^2)=log_5(9)#

#color(white)("XXX")rarr x^2=9#

#color(white)("XXX")rarr x=3# (since the argument of #log# can not be negative)

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