Logarithm Question

Question

log base 4 of (2X) + log base 4 of (X+7)=2

 

Do I need to check for extraneous?

Why doesn't the answer -8 work?

Answer 1

From the Product Identity of logarithms we know the following:

 

• Log_b (x) + Log_b (z) = Log_b (x*z)  Apply this to the given equation.
• Log₄ (2x) + Log₄ (x + 7) = Log₄ ((2x)(x + 7)) = 2
• Log4 ((2x)(x + 7)) = 2  From this equation raise each side of the equation to the 4th power to eliminate the Log.
• (2x)(x +7) = 4^2
• 2x^2 + 14x - 16 = 0 - Equation 1, Factor this quadratic equation by decomposition because the value of the coefficient (a) is greater than 1.
• a = 2, b = 14, c = -16. Use (m) and (n) as the unknowns.
• m + n = b = 14
• m(n) = a*c = 2(-16) = -32. You get 16, -2. Now decompose the 14x term in Equation 2.
• 2x^2 + (16 - 2)x - 16 = 0
• 2x^2 + 16x -2x -16 = 0.  Group common terms.
• (2x^2 + 16x) + (-2x - 16) = 0. Now use GCF.
• 2x (x - 8) - 2(x + 8) = 0. Factor again.
• (x+8) (2x -2) = 0
• x + 8 = 0     or  2x - 2 = 0
• x =-8          or    2x = 2 which implies that x = 1.

There is no Log of a negative number so x = 1 is the answer.

Answer 2

log₄(2x) + log₄(x+7) = 2

 

log₄[(2x)(x+7)] = 2

 

(2x)(x+7) = 4²

 

2x² + 14x - 16 = 0

 

x² + 7x - 8 = 0

 

(x+8)(x-1) = 0

 

x = -8 or x = 1

 

If x = -8, then the original equation becomes log₄(-16) + log₄(-1) = 2.

Since the log of a negative number is undefined, -8 is not a solution after all.  

 

So, the only solution is x = 1.

 

 

 

Answer 3

2x(x+7)=16

 

2x^2+14x-16=0

 

x^2+7x-8=0

 

(x-1)(x+8)=0

 

x=1. Or x=-8

 

But x=-8 is not in the domain of log(x+7)

 

so, the only answer is x=1