# log x + log 2x = 3 log 2 find x values

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log x + log 2x = 3 log 2

We know that:

log a + log b = log a*b

==> log x + log 2x = log 2x^2

==> log 2x^2 = 3 log 2

Also, we know that:

a log b = log b^a

==> 3 log 2 = log 2^3 = log 8

==> log 2x^2 = log 8

==> 2x^2 = 8

Now divide by 2:

==> x^2 = 4

**==> x= 2**

log x+log2x=3log2

log(x*2x)=log2 to the power 3

log2x^2=log8

x^2=8/2

x=+_2

logx+log2x= 3log3.

To solve for x.

Solution:

We use the following property of logarithms:

loga+logb = logab...(1)

And m*loga = loga^m...(2)

So logx+log2x= 3log2.Using property (1), thegiven equation becomes logx*3x = 3log2

log3x^2 = log2^3. Here we used the property (2).

Taking antilog, we get:

3x^2 = 2^3 = 8

x^2 = 8/3.

x = sqrt(8/3). Or x= -sqrt(8/3), but the negative solution is invalid as logarithms are undefined for negatives .

We'll start by imposing constraints of existence of logarithms:

x>0

2x>0

The interval of admissible values of x is (0 , +infinite).

Now, we'll solve the equation. We'll start by applying the power property of logarithms:

3 log 2 = log 2^3 = log 8

Now, we'll subtract log 2x both sides:

log x = log 8 - log 2x

We'll apply the quotient rule of logarithms:

log a - log b = log a/b

log x = log 8/2x

log x = log 4/x

Since the bases are matching, we'll apply the one to one property:

x = 4/x

x^2 = 4

x1 = +sqrt4

x1 = +2

x2 = -2

We'll reject the second solution because it doesn't belong to the interval of admissible values.

**The only solution of the equation is x = 2.**