Question

log₂ x + 1÷2 log₂(x+2)=2​

Answer 1

GIVEN :–

$\\ \bf \implies log_{2}(x) + \dfrac{1}{2} log_{2}(x + 2) = 2$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \bf \implies log_{2}(x) + \dfrac{1}{2} log_{2}(x + 2) = 2$

$\\ \bf \implies log_{2}(x) + log_{2}( \sqrt{x + 2} \: ) = 2$

$\\ \bf \implies log_{2}(x \sqrt{x + 2} \: ) = 2$

$\\ \bf \implies x \sqrt{x + 2} \: = {(2)}^{2}$

$\\ \bf \implies x \sqrt{x + 2} \: = 4$

$\\ \bf \implies x^{2}(x + 2)\: = 16$

$\\ \bf \implies x^{3}+ 2 {x}^{2} - 16 = 0$

• We should write this as –

$\\ \bf \implies (x - 2)( {x}^{2} + 4x + 8)= 0$

$\\ \bf \bigg[ \: \because \: {x}^{2} + 4x + 8 > 0\:\:because\:\: D<0\bigg]$

• So –

$\\ \bf \implies (x - 2)= 0$

$\\ \bf \implies \large\red{\boxed{ \bf x = 2}}$

• Hence , The value of 'x' is 2.

Answer 2

Step-by-step explanation:

question :

• log₂ x + 1÷2 log₂(x+2)=2

to find :

• find log = ?

• find log = ?

solution :

• log2 (x) + 1/2 log2 ( x+ 2 ) = 2

• log2 (x) + log2 (√x + 2) = 2

• log2 (x√x + 2 ) = 2

X√x + 2 = 2

X√x + 2 = 4

x(x+2) = 16

x + 2x - 16 = 0

(x − 2)(x + 4x + 8) = 0

x + 4x + 8 >0

the answer we get = (x - 2) = 0

the answer is = 2