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How do you solve #log_b9+log_bx^2=log_bx#?

1 Answer

Jul 25, 2016

#x=1/9#

Explanation:

Since #loga+logb=log(ab)#, you can write

#log_b 9x^2=log_b x#

that's equivalent to:

#9x^2=x and x>0#

and the solution is:

#x=1/9#

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