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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: solve for x log4(x-6)+log4x=2 log base 4 (x-6)+log base 4 x=2      Log On


   



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) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
.
solve for x
+log%284%2C%28x-6%29%29%2Blog%284%2Cx%29=2+
.
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
.
+%28x-6%29%28x%29+ = +x%5E2+-+6x+
.
The new log would be
.
+log%284%2C%28x%5E2+-+6x%29%29=2+
.
From here we will change this logarthimic equation into an exponential equation
.
This is how you change a logarthmic equation ( logs are a power of a base )
.
+log+%28+3%2C+243+%29+=+5+ = +3%5E5+=+243+
.
For our equation
.
+log%284%2C%28x%5E2+-+6x%29%29=2+ = +4%5E2+=+x%5E2+-+6x+ = +16+=+x%5E2+-+6x+
.
Now we can move the "16" to the right and solve the quadratic equation
.
+16+=+x%5E2+-+6x+ = +16+-+16+=+x%5E2+-+6x+-+16+ = +0+=+x%5E2+-+6x+-+16+ or +x%5E2+-+6x+-+16+=+0
.
+x%5E2+-+6x+-+16+=+0, we can factor this equation,
.
( x )(x ), first name all the factors of (-16)
.
One of these pairs will have to add up to (-6)
.
(+-16+, +1+) ( added will equal (-15) )
.
(+16+, +-1+) ( added will equal "15" )
.
(+%288%29+, +-2+) ( added will equal "6")
.
(+highlight%28-8%29+, +highlight%282%29+) ( added will equal (-6))
.
You put the highlighted factors into the parentheses with the "x"'s
.
+%28+x+-+8%29%28x+%2B+2+%29+=+0+, you can check by using the foil method, I checked and it is the same as +x%5E2+-+6x+-+16+
.
You can find "x" by placing both factors equal to "0"
.
+%28+x+-+8%29+=+0+ = +x+-+8+=+0+, move the (-8) over to the right side
.
+x+-+8+=+0+ = +x+-+8+%2B+8+=+0+%2B+8+, +x+=+8+
.
or
.
+%28x+%2B+2+%29+=+0+ = +x+%2B+2++=+0+, move the "2" to the right side
.
+x+%2B+2++=+0+ = +x+%2B+2+-+2+=+0+-+2+, +x+=+%28-2%29+
.
+x+=+8+ and +x+=+%28-2%29+, 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
.
(x = 8), +log%284%2C%28x-6%29%29%2Blog%284%2Cx%29=2+ = +log%284%2C%288-6%29%29%2Blog%284%2C8%29=2+ = +log%284%2C2%29%2Blog%284%2C8%29=2+ ( doesn't make negative, so "8" is a good answer
.
(x = (-2)), +log%284%2C%28x-6%29%29%2Blog%284%2Cx%29=2+ = +log%284%2C%28-2-6%29%29%2Blog%284%2C-2%29=2+ = +log%284%2C%28-8%29%29%2Blog%284%2C-2%29=2+ ( it creates a negative number, so (-2) is no good)
.
The only answer that works is +8+, +x+=+8+
.
The graph of this equation is
.
+graph+%28+400%2C400%2C-10%2C10%2C-10%2C10%2C+log%284%2C%28x%5E2+-+6x%29%29-2%29+
.
Remember though that "x" can only equal "8", "x" can't be equal to (-2) since this answer made the logs negative
.
+x+=+8+
.
Hope I helped, Levi