Question

If 2log(x+1) = 2log2+logx , then find the value of x

Answer 1

2log(x+1)=2log2 +logx
or, log(x+1)²=log(2)² + logx
Or, log(x+1)²=log(2² * x)
Or, (x+1)² = 4x
Or, x²+2x+1 - 4x =0
Or, x²- 2x+1=0
Or, (x-1)²=0
So ... x = 1 (Ans)

Answer 2

Required solution is x = 1

Step-by-step explanation:

The given equation is

2 log(x + 1) = 2 log2 + logx

or, log {(x + 1)²} = log (2²) + logx

or, log {(x + 1)²} = log (2² * x)

or, log {(x + 1)²} = log (4 * x)

or, log {(x + 1)²} = log (4x)

or, (x + 1)² = 4x

or, x² + 2x + 1 = 4x

or, x² - 2x + 1 = 0

or, x² - x - x + 1 = 0

or, x (x - 1) - 1 (x - 1) = 0

or, (x - 1) (x - 1) = 0

This gives x = 1

Logarithm rules:

1. loga + logb = log (ab)

2. log (a^b) = b loga

3. log (a/b) = loga - logb

4. e^(loga) = a