# What is answer of equation log base 2(x+1)=1+2log base2 (x)? a) 2 b) 3 c) 1

*print*Print*list*Cite

You need to evaluate the solution to the given equation, such that:

`log_2 (x + 1) = 1 + 2log_2 x`

Using the ower property of logarithm yields:

`log_2 (x + 1) = 1 + log_2 x^2`

Replacing `log_2 2` for 1 yields:

`log_2 (x + 1) = log_2 2 + log_2 x^2`

Converting the sum of logarithms into the logarithm of a product, yields:

`log_2 (x + 1) = log_2 (2x^2)`

Equating the arguments yields:

`x + 1 = 2x^2 => 2x^2 - x - 1 = 0`

Using quadratic formula, yields:

`x_(1,2) = (1+-sqrt(1 + 8))/4 => x_(1,2) = (1+-3)/4`

`x_1 = 1; x_2 = -1/2`

**Comparing the values ` x_1 = 1; x_2 = -1/2` with the indicated values yields that you need to select the answer c) **`x = 1.`

One easy approach that can be considered an empirical method, that helps you to find the solution, would be to replace each of the given possible values in equation, such that:

`x = 2 => log_2 (2 + 1) = 1 + 2log_2 2 => log_2 3 = 1 + 2 => log_2 3 = 3 => 3 = 2^3 => 3 = 8` invalid

Hence, you may eliminate the value `x = 2` .

`x = 3 => log_2 (3 + 1) = 1 + 2log_2 3 => log_2 4 = 1 + log_2 3^2`

`2 = 1 + log_2 9` invalid

Hence, you may eliminate the value `x = 3` .

`x = 1 => log_2 (1 + 1) = 1 + 2log_2 1 => log_2 2 = 1 + 2*0`

`1 = 1` valid

**Hence, the solution to the given equation is **`x = 1. `