What is the answer to log_{6}(4)+log_{6}(3)-log_{6}(2)=x ?

The answer to log_{6}(4)+log_{6}(3)-log_{6}(2)=x is x=1

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log_{6}(4)+log_{6}(3)-log_{6}(2)=x

lo g_{6} (4) + lo g_{6} (3) - lo g_{6} (2) = x

x = 1

Solution steps

lo g_{6} (4) + lo g_{6} (3) - lo g_{6} (2) = x

Switch sides

x = lo g_{6} (4) + lo g_{6} (3) - lo g_{6} (2)

Simplify

lo g_{6} (4) + lo g_{6} (3) - lo g_{6} (2) : 1

x = 1

3 y^{2} - 4 y - 2 = 0

2 - 2 i

9 < 3 (- \frac{1}{3} p + 6)

\frac{468}{7} - 67

What is the answer to log_{6}(4)+log_{6}(3)-log_{6}(2)=x ?
The answer to log_{6}(4)+log_{6}(3)-log_{6}(2)=x is x=1
Solve for x: log_{6}(4)+log_{6}(3)-log_{6}(2)=x
The solution is x=1