# log 2x - log (x-5) = log 5log 2x - log (x-5) = log 5

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log 2x - log (x-5) = log 5

We know that:

log a - log b = log a/b

==> log (2x/(x-5)= log 5

==> 2x/(x-5) = 5

Now cross multiply:

==> 2x = 5(x-5)

==> 2x = 5x - 25

==> 3x = 25

==> x = 25/3

We'll set the constraints of existence of logarithms:

x-5>0

x>5

2x>0

x>0

The common interval of admissible values for x is (5 ,+inf.).

Now,we'll could solve the equation in this way, also:

We'll add log (x-5) both sides:

log 2x = log 5 - log (x-5)

We'll apply the quotient property of logarithms:

log 2x = log [5/(x-5)]

Because the bases of logarithms are matching, we'll apply the one to one property:

2x = 5/(x-5)

We'll cross multiply;

2x(x-5) = 5

We'll remove the brackets:

2x^2 - 10x - 5 = 0

We'll apply the quadratic formula:

x1 = [10+sqrt(100-40)]/4

x1 = (10+2sqrt15)/4 = 4.435 < 5

x2 = (10-2sqrt15)/4 = 0.56 < 5

**Since the values of x1 and x2 do not belong to the interval of admissible values, the equation has no valid solutions.**

log2x - log(x-5) = log 5.

To solve for x:

Solution:

We use the property of logarithms: loga - logb = log (a/b) and rewrite the given equation:

log(2x/(x-5)) = log 5.

2x/(x-5) = 5 , as loga = logb omplies a = b.

2x= 5(x-5)

2x=5x-25

25 = 5x-2x =3x

x = 25/3.

Given:

log 2x - log (x - 5) = log 5

We know:

log a - log b = log a/b

Therefore:

log 2x - log (x - 5) = log [2x/(x - 5)]

Substituting this value of log 2x - log (x - 5) in the given equation:

log [2x/(x - 5)] = log 5

Taking anti-log of both sides of the equation:

2x/(x - 5) = 5

==> 2x = 5(x - 5)

==> 2x = 5x - 25

== 2x - 5x = - 25

-3x = - 25

Therefore:

x = 25/3

Here we use the relation log a-log b=log(a/b)

log 2x-log(x-5)=log[2x/(x-5)]=log5

taking antilog of both the sides: 2x/(x-5)=5

=>2x=5x-25

=>3x=25

=>x=25/3