How do you solve #-2log_5(7x)=2#?

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Answer 1 · 2016-07-27T10:26:20

#x=1/35#

You can divide all terms by 2 and have:

#log_5(7x)=-1#

you can note that

#-1=log_5(1/5)#

so you can substitute:

#log_5(7x)=log_5(1/5)#

that's equivalent to:

#7x=1/5 and x>0#

#x=1/35#