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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

 

 

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