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.
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