What is the answer to log_{10}(x+1)-log_{10}(x)=2 ?

The answer to log_{10}(x+1)-log_{10}(x)=2 is x= 1/99

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log_{10}(x+1)-log_{10}(x)=2

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

x = \frac{1}{99}

Decimal Notation

x = 0.01010 \dots

Solution steps

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

Add

lo g_{10} (x)

to both sides

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

Simplify

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

Apply log rules

x + 1 = 100 x

Solve

x + 1 = 100 x : x = \frac{1}{99}

x = \frac{1}{99}

Verify Solutions:

x = \frac{1}{99}

True

The solution is

x = \frac{1}{99}

lo g_{10} (x) = 0.5

lo g_{x} (\frac{1}{8}) = - 3

lo g_{3} (x + 7) - lo g_{3} (x - 1) = 1

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

lo g_{10} (2 x - 4) = 2

What is the answer to log_{10}(x+1)-log_{10}(x)=2 ?
The answer to log_{10}(x+1)-log_{10}(x)=2 is x= 1/99
Solve for x: log_{10}(x+1)-log_{10}(x)=2
The solution is x= 1/99