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

The answer to log_{2}(6-x)-log_{2}(4-x)=2 is x= 10/3

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

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

x = \frac{10}{3}

Decimal Notation

x = 3.33333 \dots

Solution steps

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

Add

lo g_{2} (4 - x)

to both sides

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

Simplify

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

Apply log rules

6 - x = 4 (4 - x)

Solve

6 - x = 4 (4 - x) : x = \frac{10}{3}

x = \frac{10}{3}

Verify Solutions:

x = \frac{10}{3}

True

The solution is

x = \frac{10}{3}

lo g_{5} (x - 4) = 1 - lo g_{5} (x - 8)

ln (x + 2) + ln (x) = ln (6 x) - ln (6)

ln (x^{2}) = 0

ln (2 x + 3) = 1

lo g_{b} (3^{b}) = \frac{b}{2}

What is the answer to log_{2}(6-x)-log_{2}(4-x)=2 ?
The answer to log_{2}(6-x)-log_{2}(4-x)=2 is x= 10/3
Solve for x: log_{2}(6-x)-log_{2}(4-x)=2
The solution is x= 10/3