Question

Solve for x:
$\log_{2} x + \frac{1}{2} \log_{2} (x+2)=2$

Answer 1

Answer:

2

Step-by-step explanation:

Hi,  

If y = aˣ, then we define x as logₐy  (or)

If x = logₐy , then y = aˣ

Here , a > 0 and a ≠ 1 , a is called the base of the  

logarithm.

We will be using the following properties of  

logarithm:

Additive Property : logₐx + logₐy = logₐ(xy) ,

Exponent Property : nlogₐx = logₐxⁿ

Given that log₂x + 1/2*log₂(x + 2) = 2

Multiplying by 2 on both sides, we get

2log₂x + log₂(x + 2) = 4

2log₂x = log₂x² [Using Exponent Property],

So, log₂x² + log₂(x + 2) = 4

log₂(x²)*(x + 2) = 4[Using Additive Property]

Using, the definition of logarithm, we can rewrite in

exponent form as

(x²)*(x + 2) = 2⁴ = 16

x³ + 2x² - 16 = 0

(x - 2)(x² + 4x + 8) = 0

x = 2 or x² + 4x + 8 = 0

But x² + 4x + 8 = (x + 2)² + 4 > 0

Hence, x = 2 is only permissible value.

Hope, it helps !

Answer 2

Answer:

Step-by-step explanation:

Given

$log_{2}x + \frac{1}{2} log_{2}( x + 2) = 2\\\\\implies log_{2}x + log_{2}\sqrt{x+2 } = 2\\\\\implies log_{2} x\sqrt{x+2} = 2\\\\\\\implies x\sqrt{x+2} = 2^{2}\\\\\textsf { Square both the sides }\\\\x^{2}( x+2) = 16\\\\\implies x^{3}+2x^{2}=16\\\\\implies x^{3}+2x^2-16 = 0\\\\\implies x^{3} - 2x^2 + 4x^2 - 16 =0\\\\\implies x^2(x-2) + 4( x^2 - 2^2) = 0\\ \\\implies x^2(x-2) + 4(x+2)(x-2)=0\\\\\implies (x -2 ) [ x^2 + 4( x + 2 ) ] = 0\\\\ x - 2 = or  x^2 + 4x + 2 = 0 \\\\x = 2$