log x + log(2x-8) =1

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logx + log(2x-8) =1

We know that:

logx + logy= log xy

and log 10 = 1

==> logx(2x-8) = log 10

==> log (2x^2 -8x ) = log 10

==> 2x^2 -8x = 10

Divide by 2 :

==> x^2 -4x =5

Now move 5 to the left side:

==> x^2 -4x -5=0

Now factorize:

==> (x-5)(x+1) =0

==> x1= 5

==> x2= -1 ( this is impossible because log (-1) is not defined)

Then the only solution is:

x= 5

To check:

log5 + log (2(5) -8) =1

log5 + log 2 = 1

log (5*2) = 1

log 10 =1

1=1

To solve logx+log(2x-8) = 1.

Solution:

We use loga +logb = logab. And 1 log (base k) k .

Therefore logx+log(2x+8) = 1 becomes:

logx(2x+8) = log 10. Taking antilog,

x(2x+8) = 10

2x^2+8x -10 = 0

2x^2+10x-2x-10 = 0

2x(x+5) -2(x+5) = 0

(x+5)(2x-2) = 0.

(x+5) = 0, 2x-2 = 0

x+5=0 gives x =-5 which makes log imaginary but still a valid .

2x-2 = 0 gives 2x=2, x =1 . lograrithmatic equation remains real when x =1.

Given:

log x + log(2x -8)

We know:

log A + log B = log (A*B)

and that:

log 10 = 1

Therefore the given equation can be simplified to:

log [x(2x -8)] = log 10

Therefore:

x(2x -8) = 10

2x^2 - 8x = 10

Dividing the above equation by 2, and transferring all the terms on left hand side we get:

x^2 - 4x - 5 = 0

(x^2 - 4x + 4) - 9 = 0

(x - 2)^2 - 3^2 = 0

(x - 2 + 3)(x - 2 - 3) = 0

(x + 1)(x - 5) = 0

Therefore:

X = -1, and 5

But as log(-1) is undefined, x = 5 is the only acceptable value.

Answer:

x = 5

First, we'll discuss the constraints of existence of logarithms:

x>0 and 2x-8>0

2x>8

x>4

So, for both logarithms to exist, the values of x have to be in the interval (4,+inf.).

Now, we'll solve the equation, using the product property of logarithms: the sum of logarithms is the logarithm of the product.

log x + log(2x-8) = log [x(2x-8)]

The equation will become:

log [x(2x-8)] = 1

But 1 = log 10

We'll re-write the equation:

log [x(2x-8)] = log 10

We'll use the one to one property:

[x(2x-8)] = 10

We'll opent the brackets:

2x^2 - 8x - 10 = 0

We'll divide by 2:

x^2 - 4x - 5 = 0

We'll apply the quadratic formula:

x1 = [4+sqrt(16+20)]/2

x1 = (4+6)/2

x1 = 5

x2 = (4-6)/2

x2 = -2/2

x2 = -1

**Since the second value of x is not in the interval of convenient values, the equation will have only a solution, namely x = 5.**

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