Log 3x^3-log x^2= log 27 - log x​

First of all, this equation is defined on ]3,+∞[ because you need x+3>0 and x−3>0 at the same time or the log won't be defined.

The log function maps a sum into a product, hence log(x+3)+log(x−3)=27⇔log[(x+3)(x−3)]=log27.

You now apply the exponential function on both sides of the equation : log[(x+3)(x−3)]=log27⇔(x+3)(x−3)=27⇔x2−9=27⇔x2−36=30. This is a quadratic equation that has 2 real roots because Δ=−4⋅(−36)=144>0

You know apply the quadratic formula x=−b±√Δ2a with a=1 and b=0, hence the 2 solutions of this equation : x=±6

−6∉]3,+∞[ so we can't keep this one. The only solution is x=6.

Step-by-step explanation:

I THINK IT HELP U

a

b

d

step-by-step explanation: