Question

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

Accepted Answer

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

Answer

The value of x has to be determined given that  log(2x) + log(x+5) = 2.
The base of the logarithm is 10. The logarithm of a number x to base b is defined such that if log_b x = y, x = b^y. Also, the sum log a+ log b = log a*b
Now the given equation log(2x) + log(x+5) = 2 can be rewritten as:
log(2x*(x+5)) = 2
2x*(x +5) = 10^2
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 - 5)(x + 10) = 0
x = 5 and x = -10
But the logarithm of a negative number is not defined, therefore we can eliminate the solution x = -10
The solution of the equation  log(2x) + log(x+5) = 2 is x = 5

Answer

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.

Answer

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

Answer

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.

Math

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

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