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

The answer to log_{2}(x+1)-log_{2}(x)=4 is x= 1/15

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

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

x = \frac{1}{15}

Solution steps

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

Add

lo g_{2} (x)

to both sides

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

Simplify

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

Apply log rules

x + 1 = 16 x

Solve

x + 1 = 16 x : x = \frac{1}{15}

x = \frac{1}{15}

Verify Solutions:

x = \frac{1}{15}

True

The solution is

x = \frac{1}{15}

lo g_{9} (b + 7) - 9 = - 8

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

lo g_{4} (3 x) = 2.2

1 + \frac{1}{x} - 2 \cdot ln (x) = 0

x, ln (x) = - 0.123

What is the answer to log_{2}(x+1)-log_{2}(x)=4 ?
The answer to log_{2}(x+1)-log_{2}(x)=4 is x= 1/15
Solve for x: log_{2}(x+1)-log_{2}(x)=4
The solution is x= 1/15