logarithmsWhat is x if (log2 x)^2+log2 (4x)=4?
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We need to solve (log(2) x)^2 + log(2) (4x) = 4
Use the property that log a*b = log a + log b
(log(2) x)^2 + log(2) (4x) = 4
=> (log(2) x)^2 + log(2) 4 + log(2) x = 4
=> (log(2) x)^2 + 2 + log(2) x = 4
=> (log(2) x)^2 + log(2) x = 2
Let log(2)x = y
=> y^2 + y - 2 = 0
=> y^2 + 2y - y - 2 = 0
=> y(y + 2) - 1(y + 2) = 0
=> (y - 1)(y + 2) = 0
=> y = 1 and y = -2
log(2) x = 1 => x = 2
log(2) x = -2 => x = 1/4
The solutions for x are 2 and 1/4
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First, we'll use the product rule of logarithms and we'll re-write the term log2 (4x).
log2 (4x) = log2 4 + log2 x
log2 (4x) = log2 2^2 + log2 x
We'll apply the power rule of logarithms:
log2 (4x) = 2log2 2 + log2 x
But log2 2 = 1
log2 (4x) = 2 + log2 x
We'll substitute the term log2 (4x) in the given equation:
(log2 x)^2 + 2 + log2 x = 4
We'll substitute log2 x = t
We'll re-write the equation in t:
t^2 + 2 + t - 4 = 0
We'll combine like terms:
t^2 + t - 2 = 0
We'll apply the quadratic formula:
t1 = [-1+sqrt(1 + 8)]/2
t1 = (-1+3)/2
t1 = 1
t2 = (-1-3)/2
t2 = -2
We'll put:
log2 x = t1
log2 x = 1
x = 2^1
x = 2
log2 x = t2
log2 x = -2
x = 2^-2
x = 1/2^2
x = 1/4
Since both solutions are positive, we'll accept them: {1/4 ; 2}.
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