Question

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

Answer 1

`x=(-6+sqrt72)/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 (x) + log_2 (x+6) = 3`

`log_2 (x(x+6)) = 3`

`log_2 (x^2 +6x) = 3`

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

`2^(log_2) (x^2 + 6x) = 2^3` (`2^(log_2)` cancels out)

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

`x^2 + 6x - 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=(-6+sqrt72)/2` and `x=(-6-sqrt72)/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=(-6+sqrt72)/2` works)