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

The answer to log_{4}(4x)-log_{4}(2x-4)=3 is x= 64/31

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

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

x = \frac{64}{31}

Solution steps

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

Add

lo g_{4} (2 x - 4)

to both sides

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

Simplify

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

Apply log rules

4 x = 64 (2 x - 4)

Solve

4 x = 64 (2 x - 4) : x = \frac{64}{31}

x = \frac{64}{31}

Verify Solutions:

x = \frac{64}{31}

True

The solution is

x = \frac{64}{31}

lo g_{5} (x) = 1

lo g_{\frac{2}{3}} (x) = - 2

lo g_{10} (7 x) = 2

ln (x + 1) = 3

ln (x + 3) = 2

What is the answer to log_{4}(4x)-log_{4}(2x-4)=3 ?
The answer to log_{4}(4x)-log_{4}(2x-4)=3 is x= 64/31
Solve for x: log_{4}(4x)-log_{4}(2x-4)=3
The solution is x= 64/31