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

4 Answers | Add Yours

hala718's profile pic

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

Posted on

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's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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’ve answered 318,925 questions. We can answer yours, too.

Ask a question