What is the answer to log_{2}(x+1)-log_{2}(x-4)=3 ?

The answer to log_{2}(x+1)-log_{2}(x-4)=3 is x= 33/7

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log_{2}(x+1)-log_{2}(x-4)=3

lo g_{2} (x + 1) - lo g_{2} (x - 4) = 3

x = \frac{33}{7}

Solution steps

lo g_{2} (x + 1) - lo g_{2} (x - 4) = 3

Add

lo g_{2} (x - 4)

to both sides

lo g_{2} (x + 1) - lo g_{2} (x - 4) + lo g_{2} (x - 4) = 3 + lo g_{2} (x - 4)

Simplify

lo g_{2} (x + 1) = 3 + lo g_{2} (x - 4)

Apply log rules

x + 1 = 8 (x - 4)

Solve

x + 1 = 8 (x - 4) : x = \frac{33}{7}

x = \frac{33}{7}

Verify Solutions:

x = \frac{33}{7}

True

The solution is

x = \frac{33}{7}

(lo g_{4} (1 + x^{2}))^{2} = 0.25

lo g_{5} (x + 3) = 1

lo g_{3} (x + 2) = lo g_{3} (5)

ln (x - 1) = 0

lo g_{10} (x + 1) - lo g_{10} (x) = 2

What is the answer to log_{2}(x+1)-log_{2}(x-4)=3 ?
The answer to log_{2}(x+1)-log_{2}(x-4)=3 is x= 33/7
Solve for x: log_{2}(x+1)-log_{2}(x-4)=3
The solution is x= 33/7