### What is x if (log2 x)^2+log2 (4x)=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}.**

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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