log 2 (2x) + log 4 (4x^2) = 3 solve for x

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log 2 (2x) + log 4 (4x^2) = 3

First let us rewrite :

We know that:

log a b = log c b/log c a

==> log 4 (4x^2) = log 2 (4x^2) / log 2 4

But log 2 4 = 2

==> log 4 (4x^2) =(1/2) log 2 (4x^2) = log 2 (4x^2)^1/2

==> log 2 (2x) + log 2 (4x^2)^1/2 = 3

==> log 2 (2x) + log 2 (2x) = 3

We know thatL log a + log b = log a*b

==> log 2 (2x)*(2x) = 3

==> log 2 (4x^2) = 3

==> 4x^2 = 2^3

==> 4x^2 = 8

==> x^2 = 2

==> **x = sqrt2**

log2 (2x) +log4 (4x^2) = 3.

To solve for x.

Solution:

We see that there are 2 bases of logarithms. We make a common base of loggarithms .

log b (c) = log a((c)/ loga (b).

So the 2nd term log4 (4x^2) = log2 (4x^2)/ log2 (4) = {log2(4x^2)}/2 , as log2 (4) = log2 (2^2) = 2 log2 (2) = 2*1 =2.

So with the above , given equation becomes:

log2 ((2x) + {log2 (4x^2)}/2 = 3 = log2 (2^3). Multiply by 2.

2 log2 (2x) + log2 (4x^2) =2* log 2 (8)

log2{ (2x)^2} +log2 (4x^2) = 2*log2 (8)

log2 {(4x^2)(4x^2)} = log2 (64)., as log a+logb = logab.

16x^4 = 64, by one to one property.

x^4 =64/16 = 4

x = (4)^(1/4) = +or- 2^(1/2) Or (-2)^(1/2).

x = 2^(1/4) or x = -2^(1/4) or x = [2^(1/2)]i or x = [-2^(1/2)]i

The first two are real solutions and the last two are imaginary solutions.

We notice that the bases of logarithms are not matching, so, we'll choose as common base 2.

We'll change the base of the second logarithm log 4 (4x^2):

log 2 (4x^2) = log 4 (4x^2) * log 2 4

But log 2 4 = log 2 2^2 = 2*log 2 2 = 2*1 = 2

log 2 (4x^2) = 2*log 4 (4x^2)

So, the second term log 4 (4x^2) will be substituted by

log 2 sqrt(4x^2)

We'll rewrite the equation:

log 2 (2x) + log 2 sqrt(4x^2) = 3

log 2 (2x) + log 2 (2x) = 3

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

log 2 (2x*2x) = 3

4x^2 = 2^3

4x^2 = 8

We'll divide by 4:

x^2 = 2

x = +/-sqrt2

**We'll accept only the positive solution, x = sqrt2.**

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