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

The answer to log_{10}(x+1)+log_{10}(2x)=2 is x=(-1+sqrt(201))/2

Solve for x: log_{10}(x+1)+log_{10}(2x)=2

The solution is x=(-1+sqrt(201))/2

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

x = \frac{- 1 + 201}{2}

Solution steps

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

Apply log rules

(x + 1) \cdot 2 x = 100

Solve

(x + 1) \cdot 2 x = 100 : x = \frac{- 1 + 201}{2}, x = - \frac{1 + 201}{2}

x = \frac{- 1 + 201}{2}, x = - \frac{1 + 201}{2}

Verify Solutions:

x = \frac{- 1 + 201}{2}

True

, x = - \frac{1 + 201}{2}

False

The solution is

x = \frac{- 1 + 201}{2}

3 ln (p) + 7 ln (1 - p) = 0

2 lo g_{e} (x) + lo g_{e} (x) = 0

lo g_{5} (P) = 10

ln (x^{3}) = 7 ln (x)

3 ln (4) = 2 ln (x)

What is the answer to log_{10}(x+1)+log_{10}(2x)=2 ?
The answer to log_{10}(x+1)+log_{10}(2x)=2 is x=(-1+sqrt(201))/2
Solve for x: log_{10}(x+1)+log_{10}(2x)=2
The solution is x=(-1+sqrt(201))/2