SOLUTION: solve for x log4(x-6)+log4x=2 log base 4 (x-6)+log base 4 x=2

Question 172610: solve for x
log4(x-6)+log4x=2
log base 4 (x-6)+log base 4 x=2
Answer by Electrified_Levi(103) (Show Source): You can put this solution on YOUR website!
Hi, Hope I can help,
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solve for x

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If you add to of the same logs together, that is the same thing as multiplying the terms together, with only one log, this is what I mean
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=
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The new log would be
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From here we will change this logarthimic equation into an exponential equation
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This is how you change a logarthmic equation ( logs are a power of a base )
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=
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For our equation
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= =
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Now we can move the "16" to the right and solve the quadratic equation
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= = or
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, we can factor this equation,
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( x )(x ), first name all the factors of (-16)
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One of these pairs will have to add up to (-6)
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(, ) ( added will equal (-15) )
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(, ) ( added will equal "15" )
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(, ) ( added will equal "6")
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(, ) ( added will equal (-6))
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You put the highlighted factors into the parentheses with the "x"'s
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, you can check by using the foil method, I checked and it is the same as
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You can find "x" by placing both factors equal to "0"
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= , move the (-8) over to the right side
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= ,
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or
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= , move the "2" to the right side
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= ,
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and , the only thing left is to put these values into the original equation, logs can't be nagative, so we need to make sure that our answers don't create negative numbers
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(x = 8), = = ( doesn't make negative, so "8" is a good answer
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(x = (-2)), = = ( it creates a negative number, so (-2) is no good)
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The only answer that works is ,
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The graph of this equation is
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Remember though that "x" can only equal "8", "x" can't be equal to (-2) since this answer made the logs negative
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Hope I helped, Levi
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