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Solve the logarithmic equation. Express your solutions in exact form only. Please show...
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High School Teacher
`log_2 (x+4) - 3 = log_2 (x+5) - log_2 (x+6)`
`log_2 (x+4) - log_2 2^3 = log_2 (x+5) - log_2 (x+6)`
`log_2 (x+4)/8 = log_2 ((x+5)/(x+6))`
`==> (x+4)/8 = (x+5)/(x+6)`
`==> (x+4)(x+6) = 8(x+5)`
`==> x^2 + 10x + 24 = 8x + 40`
`==> x^2 +2x - 16 = 0`
`==> x1= (-2+sqrt68)/2 = (-2+2sqrt17)/2 = -1+sqrt17`
`==> x2= -1-sqrt17`
But x2 is not a valid answer because the log is not defined.
==> Then the answer is : `x = -1+sqrt17` ``
Posted by hala718 on November 12, 2011 at 11:01 AM (Answer #1)
Bases of logarithms are the same. Let's move the logarithms to the left side and the number to the right side.
log 2 (x+4)+log 2 (x+6)- log2 (x+5)=3
Use the product property for the logarithms:
log 2 (x+4)+log 2 (x+6)=log 2 (x+4)(x+6)
Use the quotient property for the logarithms:
log 2 (x+4)(x+6) - log2 (x+5)= log 2 [(x+4)(x+6)/(x+5)]
Write the equation.
log 2 [(x+4)(x+6)/(x+5)]=3 =>
(x+4)(x+6) = 8(x+5)
Open the brackets
x^2 + 10x + 24 - 8x - 40 = 0
x^2 + 2x - 16=0
x1=(-2+sqrt68)/2 => x1=-1+sqrt17
Since the values of x for the logarithms to be defined is (-4,+oo), the value -1-sqrt17 is not included.
ANSWER: The equation ihas a single solution x1=-1+sqrt17.
Posted by sciencesolve on November 13, 2011 at 3:58 PM (Answer #2)
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