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

1 Answer
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Mar 11, 2018

#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)

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