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

lemjay | Certified Educator

`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`  .