find the Log?

Question

how do I find the log of:

1. Log(sub12) 1

2. Log(sub12) 12

3. Log(sub12)144

Answer 1

Question 1
0
Question 2
1
Question 3
12

Answer 2

They are 0, 1, 2 --- in that order.

That is because if y = log(sub b) x, then b^y = x. (See *** below.)

So you have:

1. y = log(sub 12) 1, i.e. 12^y = 1, which requires that y = 0;

2. z = log(sub 12) 12, i.e. 12^z = 12, therefore z = 1; and

3. w = log(sub 12) 144, i.e. 12^w = 144 (= 12^2), so w = 2.

Live long and prosper.

Note that once you're familiar with spotting powers, this question is really asking you:

What are logs to the base 12 of 12^0, 12^1, and 12^2?

--- to which the obvious answers are 0, 1, and 2.

That is how I was immediately able to write down the first line of my answer.

However, if you're NOT so familiar, the process I gave above (at ***) should help you until that time comes.

Answer 3

1. Log_12 1 = A means

12^A = 1. A =0, right?


2. Log_12 12 = A means
12^A = 12, so A = 1, right?

3. Log_12 144 = A means

12^A = 144, so what is A? It's the next number from 1.

Answer 4

1) 0 because 12^0 = 1

2) 1 because 12^1 = 12

3) 2 because 12^2 = 144

Answer 5

I will use y to equate your logs.

log_12 (1) = y

y becomes the exponent and the sub numbers become our base for all three questions.

We have:

12^y = 1

The number 12 raised to what power = 1?

Any number raised to a ZERO power = 1.

So, y = 0

++++++++++++++++++++++

log_12(12) = y

The same process all over again.

12^y = 12

When a number is raised to the power of 1, the answer is the same base.

So, 12^1 = 12.

y = 1 for question 2

+++++++++++++++++++++++++++++++

log_12(144) = y

Then 12^y = 144

What can y be this time?

Well, 12 raised to the second power = 12 times 12 = 144

So, y = 2

Answers:

For question (1), y = 0

For question (2), y = 1

For question (3), y = 2

Is this clear?

P.S. I used the letter y but you can use any letter of choice.
Another person used A, which is also okay.