Question

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

Accepted Answer

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

Answer

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

Answer

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.

Answer

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.

Math

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

Expert Answers

We’ll help your grades soar

Related Questions

Solve: log 2 (3x ) - log 4 ( x-2) = 3

x^3+2x^2-x-2=0 solve for x

log (5x-1) - log (X-2) = log 3 solve for x

Solve for x if log (x^2) - log 2x = 2.

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