# Hi! could you please help me to evaluate this problem? 〖log〗_2 (2^327-2^326 )= thank you

### 3 Answers | Add Yours

Hi, Haviv,

For this one, you need to consider that 2^372 = 2 * 2^326. So, we would have:

2^327 - 2^326 = 2 * 2^326 - 2^326

Factoring out 2^326, we get:

2^326 ( 2 - 1 ) = 2^326.

Therefore, 2^327 - 2^326 = 2^326.

So, then, we out the log = x. So:

log_2 (2^326) = x

Rewriting it as an exponential, we get:

2^x = 2^326.

So, x has to be 326.

Good luck, Haviv. I hope this helps.

Till Then,

Steve

Evaluate: `log_(2) (2^327-2^326)`

First, evaluate parenthesis. In evaluating expressions with exponents you can state that: `x^a - x^b = x^b`

This means that `2^327 - 2^326 = 2^326`

Therefore we have: `log_(2)2^326 = x`

Rewriting using the meaning of logs, we have: `log_(a)b = xhArr a^x = b`

`2^x = 2^236`

**Since we have the same base this means that**:

`x = 236`

`log_2(2^327-2^326)=log_2(2^326(2-1))`

`=log_2(2^326xx1)`

`=log_2 2^326`

`=326log_2 2`

`=326`