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

The answer to log_{4}(x)-log_{4}(x-2)=2 is x= 32/15

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

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

x = \frac{32}{15}

Solution steps

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

Add

lo g_{4} (x - 2)

to both sides

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

Simplify

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

Apply log rules

x = 16 (x - 2)

Solve

x = 16 (x - 2) : x = \frac{32}{15}

x = \frac{32}{15}

Verify Solutions:

x = \frac{32}{15}

True

The solution is

x = \frac{32}{15}

- ln (1 - p) = 0.002

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

lo g_{4} (x) - lo g_{4} (x + 9) = - 1

ln (n^{2} + 12) = ln (- 9 n - 2)

lo g_{2} (x + 1) - 2 lo g_{2} (3) = 3

What is the answer to log_{4}(x)-log_{4}(x-2)=2 ?
The answer to log_{4}(x)-log_{4}(x-2)=2 is x= 32/15
Solve for x: log_{4}(x)-log_{4}(x-2)=2
The solution is x= 32/15