# How do you solve log_2 (x) + log_2 (x+6) = 3?

Mar 11, 2018

$x = \frac{- 6 + \sqrt{72}}{2}$

#### Explanation:

When you add two logs with the same base, based on the logarithm rules, it's the same as multiplying them.

${\log}_{2} \left(x\right) + {\log}_{2} \left(x + 6\right) = 3$

${\log}_{2} \left(x \left(x + 6\right)\right) = 3$

${\log}_{2} \left({x}^{2} + 6 x\right) = 3$

To get rid of the log in order to isolate the variables:

${2}^{{\log}_{2}} \left({x}^{2} + 6 x\right) = {2}^{3}$ (${2}^{{\log}_{2}}$ cancels out)

${x}^{2} + 6 x = 9$ (subtract 9 and set equal to zero, so that you are able to factor the equation)

${x}^{2} + 6 x - 9 = 0$

Since there are no like terms, use the quadratic formula to solve:

x=(-b ± sqrt(b^2 -4ac))/(2a)

ax + bx + c=0

x=(-6 ± sqrt(-6^2 -4(1)(-9)))/(2(1))

x= (-6± sqrt(36- (-36)))/2

x= (-6±sqrt(72))/2

$x = \frac{- 6 + \sqrt{72}}{2}$ and $x = \frac{- 6 - \sqrt{72}}{2}$

Remember to check both answers to see if they work, by plugging them back into the original equation (if you do this you see that only $x = \frac{- 6 + \sqrt{72}}{2}$ works)