Log rules...?

Question

The problem is

2^log base 2 of 12

It cancels out and equals 12, or am I mistaken?

Answer 1

You are not mistaken,

let us consider

2^log base 2 of 12=y

and let log base 2 of 12=t......................(1)

the equation becomes, 2^t=y

this is exponential fn.
converting it into logarithmic function
log base 2 of y=t
log base 2 of y= log base 2 of 12 from(1)

comparing both sides y= 12

Answer 2

Correct, the answer is 12.

2^log_2 (12)

So 2 has the power of log_2 (12). Let's get the value of the power part first.
log_2 (12) = [log_10 (12)] / [log10 (2)]
Now that the equation is switched to a base 10, you can get the values handy enough for the calculator or the log tables.
log_2 (12) = [log_10 (12)] / [log10 (2)]
log_2 (12) = [1∙079 181 246] / [0∙301 029 995]
log_2 (12) = 3∙584 962 501 So this is the value of the power of the 2.
2^3∙584 962 501 = 12

Answer 3

Take log base 2 , of both sides:

it will be
log base2 of 12 times log base 2 of 2=log base2 of x

therefore x=12

you are right

P.S: BE CAREFUL FOR THE BASES
if we take the log of both sides base different from 2
the value of x will not be 12 !!!

Answer 4

Let y = 2^log{2}(12)
log{2}(y) = log{2}(12)log{2}(2) = log{2}(12)
y = 12
2^log{2}(12) = 12
Yes, it cancels out.

Answer 5

YES you are right

http://en.wikipedia.org/wiki/Logarithmic_identities#Canceling_exponentials