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

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The equation `log_2(2x) + log_4 (4x^2) = 3` has to be solved for x.

First let us bring all the logarithm to the same base, use 2 and the formula `log_b c = (log_x c)/(log_x b)`

This gives:

`log_2(2x) + log_4 (4x^2) = 3`

`log_2(2x) + (log_2 (4x^2))/(log_2 4) = 3`

Now `log_2 4 = log_2 2^2 = 2*log_2 = 2`

`log_2(2x) + (log_2 (4x^2))/2 = 3`

`2*log_2(2x) + log_2 (4x^2) = 6`

Use the formula a*log b = log b^a and log a + log b = log a*b

`log_2(4x^2) + log_2 (4x^2) = 6`

`log_2(16x^4) = 6`

`log_2 16 + log_2 x^4 = 6`

`log_2 2^4 + log_2 x^4 = 6`

`4 + log_2x^4 = 6`

`log_2 x^4 = 2`

`x^4 = 2^2`

`x^4 = 4`

`x = sqrt 2`

The solution of the equation is `x = sqrt 2`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

neela | High School Teacher | (Level 3) Valedictorian

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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.