SOLUTION: Solve for x log(base 4) x + log(base 4) (x-2) = 1

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Question 1015407: Solve for x
log(base 4) x + log(base 4) (x-2) = 1

Found 3 solutions by josmiceli, josgarithmetic, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
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use the substitution:
+1+=+log%284%2C4%29+
+log%28+4%2Cx+%29+%2B+log%28+4%2C+x-2+%29+=+log%28+4%2C4+%29+
+log%28+4%2C+x%2A%28+x-2+%29+%29+=+log%28+4%2C4+%29+
+x%2A%28+x-2+%29+=+4+
+x%5E2+-+2x+=+4+
Complete the square
+x%5E2+-+2x+%2B+%28-2%2F2+%29%5E2+=+4+%2B+%28-2%2F2+%29%5E2+
+x%5E2+-+2x+%2B+1+=+4+%2B+1+
+%28+x+-+1+%29%5E2+=+5+
Take the square root of both sides
+x+-+1+=+sqrt%285%29+
+x+=+1+%2B+sqrt%285%29+
and also:
+x+-+1+=+-sqrt%285%29+
+x+=+1+-+sqrt%285%29+
------------------
check:
+x%2A%28+x-2+%29+=+4+
+%28+1+%2B+sqrt%285%29+%29%2A%28+-1+%2B+sqrt%285%29+%29+=+4+
+-1+%2B+5+=+4+
+4+=+4+
OK
You can check the other answer

Answer by josgarithmetic(39623) About Me  (Show Source):
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Pure text , log(4,x)+log(4,(x-2))=1

Adding the triple-braces rendering tags,
log%284%2Cx%29%2Blog%284%2C%28x-2%29%29=1

log%284%2C%28x%28x-2%29%29%29=1
log%284%2C%28x%5E2-2x%29%29=1
4%5E1=x%5E2-2x
x%5E2-2x=4
x%5E2-2x-4=0----not factorable

x=%282%2B-+sqrt%28%28-2%29%5E2-4%2A1%2A%28-4%29%29%29%2F2
x=%282%2B-+sqrt%2820%29%29%2F2
x=%282%2B-+2%2Asqrt%285%29%29%2F2
highlight%28x=1%2B-+sqrt%285%29%29

Answer by ikleyn(52834) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve for x
log(base 4) x + log(base 4) (x-2) = 1
-----------------------------------------

It is equivalent to

x*(x-2) = 4,     (1)

x%5E2+-2x+-4 = 0,

x%5B1%2C2%5D = %282+%2B-+sqrt%284+%2B4%2A4%29%29%2F2 = 2+%2B-+sqrt%2820%29%29%2F2 = 1+%2B-+sqrt%285%29.

So, the solutions of the equation (1) are 1%2Bsqrt%285%29 and 1-sqrt%285%29.

Only positive value x = 1+%2B+sqrt%285%29 satisfies the original equation.