# Logarithmic Equation `log_6 3 + log_6 (2x^2 +4) = 2`

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`log_6 3+log_6(2x^2+4)=2`

Now we use the fact that sum of logarithms is logarithm of product i.e.

`log_b x+log_b y=log_b xy`

Hence we get

`log_6(3(2x^2+4))=2`

`6x^2+12=6^2`

`6x^2=36-12`

`6x^2=24`

`x^2=4`

`x_(1,2)=pm2`

**So your solutions are** `x_1=-2` **and** `x_2=2.`