# log 2x + log 5 = log (4x+2)log 2x + log 5 = log (4x+2)

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### 5 Answers

You may approach different methods to solve this equation, hence, either you may convert the sum into a product, or you move the terms that contain x to the left side and the constant term to the right.

Converting the sum in a product yields:

Hence, substituting for yields:

Equating the numbers yields:

**Hence, evaluating the solution to the given equation yields **

log 2x + log 5 = log (4x + 2)

We know that:

log a + log b = log a*b

==> log 2x*5 = log (4x+2)

==> log 10x = log (4x+2)

==> 10 x = 4x + 2

==> 6x = 2

==> x= 2/6 = 1/3

==> x= 1/3

We'll impose the constraints of existence of logarithms:

2x>0

x>0

4x+2>0

4x>-2

x>-1/2

The common interval of admissible values for x is: (0,+inf.).

Now, we'll solve the equation.

We'll subtract log 5 both sides:

log 2x + log 5 = log (4x+2)

log 2x = log (4x+2) - log 5

We'll use the quotient property of logarithms:

log 2x = log [(4x+2)/5]

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

2x = [(4x+2)/5]

10x = 4x+2

We'll isolate x to the left side. For this reason, we'll subtract 4x both sides:

10x - 4x = 2

6x = 2

x = 2/6

**x = 1/3 > 0**

**Since the value is positive and belongs to the interval of admissible values, it is a valid solution.**

log2x +log5 = log(4x+2). To solve for x.

Solution:

We use log ab = loga+logb.

log2x = log2+logx.

log(4x+2) = log2(2x+1) = log2 +log(2x+1)

So the equation becomes:

log2 +logx +log5 = log(4x+2) = log 2*(2x+1) = log2+log(2x+1)

logx +log5 = log(2x+1)

log5x = log(2x+1). Take antilogarithms:

5x= 2x+1

5x-2x = 1

3x= 1. Divide by 3.

x = 1/3.

Given:

log 2x + log 5 = log (4x + 2)

We know:

log a + log b = log ab

Therefore:

log 2x + log 5 = log (2x*5) = log 10x

Thus the given equation can be simplified as:

log 10x = log (4x + 2)

Taking anti log of both the sides:

10x = 4x + 2

==> 10x - 4x = 2

==> 6x = 2

Dividing both sides by 6:

x = 2/6 = 1/3