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How do you solve #2log x = 4log 7#?

1 Answer

Dec 24, 2015

#x=49#

Explanation:

Remember:
#color(white)("XXX")a*log(c) = log(c^a)#

Therefore
#color(white)("XXX")2log(x)=4log(7)#

#rarrcolor(white)("XXX")log(x^2)=log(7^4)#

#rarrcolor(white)("XXX")x^2=7^4#

#rarrcolor(white)("XXX")x=7^2=49#

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