log(2x) + log(x+5) = 2

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log(2x) = log(x+5) =2

We know that logx + logy= log(xy)

==> log(2x) + log(x+5) = log(2x(x+5)

==> log(2x^2 +10x) = 2

But log100 = 2

==> log(2x^2 + 10x) = log 100

==> 2x^2 + 10x = 100

Divide by 2:

==> x^2 + 5x = 50

Subtract 50 from both sides:

==> x^2 +5x -50=0

Factorize:

==> (x+10)(x-5) =0

==> x1= 5

==> x2= -10 ( this is impossible because the function is not defined)

**Then the solution is x= 5**

log(2x)+log(x+5) =2.

We know loga+logb = logab.So the equation becomes:

log(2x(x+5)) = 2 = log 10^2. Taking antilogarithms,

2x(x+5) =100.Divide by 2.

x(x+5) = 50

x^2+5x = 50

x^2+5x-50 = 0

x^2+10x-5x-50 = 0

x(x+10)-5(x+10) = 0

(x+10)(x-5) = 0

x+10 = 0 , x-5 = 0

x=-10. Gives the imaginary logarimatic equation but still the solution holds good.

x = 5. Real solution of real logarimatic equation.

We know:

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

We also know:

log 100 = 2

Therefore:

log(2x) = log (x + 5) = 2 = Log 100

==> log [2x(x + 5)] = Log 100

==> log (2x^2 + 10x)= Log 100

Therefore:

2x^2 + 10x = 100

==> 2x^2 + 10x - 100 = 0

==> x^2 + 5x - 50 = 0

==> x^2 + 10x - 5x - 50 = 0

==> x(x + 10) - 5(x + 10) = 0

==> (x -5)(x + 10) = 0

Therefore:

x = 5, or x = -10

Taking the value of x = -10 will result in terms in the origibla equations (log 2x, and log [x + 5]) that are undefined.

Therefore:

x = 5

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

2x>0 and x+5>0

x>-5

x>0

So, for both logarithms to exist, the values of x have to be in the interval (0,+inf.).

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

log(2x) + log(x+5) = log [2x(x+5)]

The equation will become:

log [2x(x+5)] = 2

But 1 = log 10

We'll re-write the equation:

log [2x(x+5)] = 2 log 10

log [2x(x+5)] = log 100

We'll use the one to one property:

[2x(x+5)] = 100

We'll open the brackets:

2x^2 + 10x - 100 = 0

We'll divide by 2:

x^2 + 5x - 50 = 0

We'll apply the quadratic formula:

x1 = [-5+sqrt(25+200)]/2

x1 = (-5+15)/2

x1 = 5

x2 = (-5-15)/2

x2 = -20/2

x2 = -10

**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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