# How do you solve: `log_2x + log_2(x+1) = 1`

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`log_2x+log_2 (x+1) = 1`

To solve for x, apply the product rule of logarithm which is `log_am + log_an = log_a(m*n)` .

`log_2 (x(x+1)) = 1`

`log_2(x^2+x)=1`

Then, express the equation to its equivalent exponential form.

Note that `log_a b=m` can be written as `a^m =b` .

`2^1 = x^2+x`

`2=x^2+x`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2+x-2=0`

Then, factor.

`(x+2)(x-1)=0`

Set each factor to zero and solve for x.

`x+2=0` and `x-1=0`

`x=-2` `x=1`

To check, substitute the values of x to the original equation.

`x=-2` , `log_2 (-2)+log_2(-2+1)=1`

`log_2 -2+log_2-1=1` (Invalid, because negative argument of logarithm is not allowed.)

x=1, `log_2 1+log_2(1+1) = 1`

`log_2 1+log_2 = 1`

`log_2 (2*1)=1`

`log_2 2 = 1`

`1 = 1` (True)

**Hence, the solution to the equation `log_2x + log_2(x+1)= 1` is `x=1` .**

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