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

3 Answers | Add Yours

steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

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

baxthum8's profile pic

baxthum8 | High School Teacher | (Level 3) Associate Educator

Posted on

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`

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

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

`=log_2(2^326xx1)`

`=log_2 2^326`

`=326log_2 2`

`=326`

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question