Log2(x-3)+Log2(x+2)=-4

Question

Log₂(x-3)+Log₂(x+2)=-4

 

i have X²-X-6 = 1/16    but solve for X

 

Can someone walk me through the steps please?

Answer 1

We have a product sum of log functions.  We can rewrite this as a product of the logs.

 

Log₂[(x - 3)(x + 2)] = -4

 

Log₂(x² - x - 6) = -4

 

Since -4 is the solution of the Log, it will be the exponent of the 2.

 

2^-4 = x² - x - 6

 

1/16 = x² - x - 6

 

Multiply both sides of equation by 16 to get rid of the denominator.

 

1 = 16x² - 16x - 96

 

Subtract 1 on both sides of equation.

 

0 = 16x² - 16x - 97

 

Use the quadratic formula:

 

x = (16 ± √(16² - 4(-1552))) / 32

x = (16 ± √(256 + 6208)) / 32

x = (16 ± 80.40) / 32

 

x₁ = 3.01

x₂ = -2.01

 

Answer 2

Hi Osiris,

 

You have the quadratic correct. I suggest multiplying both sides of the equation by 16 to eliminate the fraction, then set equal to zero and finish solving by a) factoring, b) completing the square, or c) quadratic formula.

 

So,

 

16x²-16x-96=1

 

16x²-16x-97=0

 

x = 16 +/- sqrt (6464)/(32)

 

x = 1/4(2+ sqrt(101))

 

rejecting negative root of the quadratic.