# How do you solve log_2x + log_2(x+1) = log26?

Oct 30, 2015

$x = 1 , 2078$

#### Explanation:

By using the laws of logs, we may rewrite this equation as

${\log}_{2} \left[x \cdot \left(x + 1\right)\right] = {\log}_{10} 26$

$\therefore {\log}_{2} \left({x}^{2} + x\right) = {\log}_{10} 26$

$\therefore {x}^{2} + x = {2}^{\log 26} = 2 , 6665481$

This is now a quadratic equation which we may use the quadratic formula to solve for:

$\therefore x = \frac{- 1 \pm \sqrt{1 - \left(4 \times 1 \times - 2 , 6665481\right)}}{2}$

$= 1 , 2078 \mathmr{and} - 2 , 2078$

But $x = - 2 , 2078 \notin D o m {\log}_{2} x$ and so $x = 1 , 2078$ is the only feasible solution.