Question

Solve: log 2  (3x )  - log 4 ( x-2) = 3

Accepted Answer

log 2 (3x) - log 4 (x-2) = 3
We know that log a b = log c b/ log c a
==> log 4 (x-2) = log 2 (x-2)/ log 2 (4) = log 2 (x-2)/2 = (1/2) log (x-2) = log (x-2)^1/2
==> loh 2 (3x) - log (x-2)^1/2 = 3
We know that log a - log b = log a/b
==> loh 2 (3x)/(x-2)^1/2 = 3
==> 3x/ (x-2)^1/2  = 2^3
==>  3x = 8*(x-2)^1/2
Now square both sides:
==> 9x^2 = 64(x-2)
==> 9x^2 = 64x - 128
==> 9x^2 - 64x + 128 = 0
==> x1= [64 + sqrt(4096 - 4608)/18     ( impossible
Then the problem has no real solution.

Answer

First, we'll impose constraints of existence of logarithms:
3x>0
x>0
x-2>0
x>2
The common interval of admissible values is (2, +inf.)
Now, we'll solve the equation.
log 2  (3x)  - log 4 (x-2) = 3
We'll add log 4 ( x-2) both sides:
log 2  (3x) = log 4 (x-2) + 3
We'll write 3 = 3*1 = 3*log 4 4 = log 4 4^3
We'll re-write the equation:
log 2  (3x) = log 4 (x-2) + log 4 4^3
We'll use the product property of logarithms:
log 2  (3x) = log 4 [(x-2)*4^3]
We'll change the base 2 of the logarithm from the left side, into the base 4:
log 4 (3x) = log 2  (3x) * log 4 2
But log 4 2 = 1/log 2 4 = 1/log 2 2^2 = 1/2
2*log 4 (3x) = log 2  (3x)
Now, we'll re-write the equation with all logarithms having matching bases:
2*log 4 (3x) = log 4 [(x-2)*4^3]
log 4 (9x^2) = log 4 [(x-2)*4^3]
We'll use the one to one property:
9x^2 = 64x - 128
We'll move all terms to one side:
9x^2 - 64x + 128 = 0
x1 = (64+sqrt512)/18
x1 = (64+16sqrt2)/18
x1 = 16(4+sqrt2)/18
x1 = (32+8sqrt2)/9
x2 = (32-8sqrt2)/9
Since both solutions are in the interval (2,+inf.), they are valid.

Answer

log2 (3x) - log4 (x-2) = 3. To find x.
Solution:
There are 2 logarithm bases 2 and 4. We convert  base2 and  4 into logarithm of base 10 .
 log2 (3x) = log 3x/log2 , the log base is 10.
log4 (x-2) =  log(x-2)/ log4 = log (x-2)log2^2 =log(x-2)/2log2
Therefore  the given equation  becomes:
log3x / log2  - log(x-2) / 2log2  = 3
2log3x / 2log2  = log(x-2) /2log2 = 3
(2log3x - log(x-2)}/(2log2) = 3
log (3x)^2 /(x-2)  = 3 *2log 2 = log2^6. Taking antilog,
9x^2/(x-2) = 2^6 = 64
9x^2 = 64(x-2)
9x^2 -64x +128 = 0
x ={ 64+or - sqrt[64^2- (9*128*4)]}/(2*9)
= {64 +or sqrt(-512)}/18
= {32 +or- sqrt128}/9
={32 +or- 8sqrt(-2)}/9
X1 = 32+(8sqrt2)i or  x2 = 32 - (8sqrt2)i  both are solutions in complex numbers.

Math

Solve: log 2 (3x ) - log 4 ( x-2) = 3

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