Another Question - power rules

[oupavios]

I had another question which should be simple for you guys. I have a value of 2 ^149 which I know from a book I'm reading is roughly 10 ^45. But what calculation needs to be done in order to arrive at the second number?

Thanks again!

 

[Deleted member 4993]

I had another question which should be simple for you guys. I have a value of 1/2 ^149 which I know from a book I'm reading is roughly 10 ^45. But what calculation needs to be done in order to arrive at the second number?

Thanks again!

That cannot be!

(1/2)^149 = 1/2 * 1/2 * 1/2 * ...... * 1/2 * 1/2 * 1/2.....[ 1/2 multiplied with itself 149 times]

every-time you multiply the product by 1/2 - the result is smaller than the previous answer. So (1/2)^149 is a vewy vewy small number (now I am going out to hunt wabbit)

 

[oupavios]

That cannot be!

(1/2)^149 = 1/2 * 1/2 * 1/2 * ...... * 1/2 * 1/2 * 1/2.....[ 1/2 multiplied with itself 149 times]

every-time you multiply the product by 1/2 - the result is smaller than the previous answer. So (1/2)^149 is a vewy vewy small number (now I am going out to hunt wabbit)

Sorry about that. The "1/2" meant '1 out of 2'. I should have just wrote 2 ^149 which I just went back and corrected in my original question.

 

[HallsofIvy]

Elmer, the original post has been edited- it is now \(\displaystyle 2^{149}\), not \(\displaystyle (1/2)^{149}\).

Oupavios, the common logarithm of 2 is about 0.3010 meaning that \(\displaystyle 2= 10^{0.3010}\). So that \(\displaystyle 2^{149}= (10^{0.3010})^{149}= 10^{149*0.310}= 10^{44.8}\).

 

[oupavios]

Elmer, the original post has been edited- it is now \(\displaystyle 2^{149}\), not \(\displaystyle (1/2)^{149}\).

Oupavios, the common logarithm of 2 is about 0.3010 meaning that \(\displaystyle 2= 10^{0.3010}\). So that \(\displaystyle 2^{149}= (10^{0.3010})^{149}= 10^{149*0.310}= 10^{44.8}\).

Excellent! Thank you, that was tremendously helpful. From that, I was able to extrapolate that to determine the common logarithm of any number from 1 to 9, all you have to do is hit "log" on you calculator, and then enter your number, and it will give you the logarithm for that number. This opened up a whole new world for me.

Thanks again!

 

[oupavios]

One Final Peice to the Puzzle (Hopefully)

Question edited:

So we know that .5 X 10 ^164 is 500000000E+163, or 5 followed by 163 zeros. What then is the easiest way to determine how many times 10 must be multiplied by itself to equal that number?

Thanks!

 

[Deleted member 4993]

Question edited:

So we know that .5 X 10 ^164 is 500000000E+163, or 5 followed by 163 zeros. What then is the easiest way to determine how many times 10 must be multiplied by itself to equal that number?

Thanks!

No - that should be 5E+163 or 5 followed by 163 zeros

5 * 10¹⁶³ = 10^0.69897* 10¹⁶³ = 10^{(0.69897+163)} = 10^163.69897 ≈ 10¹⁶⁴

 

[oupavios]

No - that should be 5E+163 or 5 followed by 163 zeros

5 * 10¹⁶³ = 10^0.69897* 10¹⁶³ = 10^{(0.69897+163)} = 10^163.69897 ≈ 10¹⁶⁴

The "500000000E+163" is just how my calculator was displaying the result and I'm sure 5E+163 is the correct way to express it. But I'm not sure if you understood the question so let me try to phrase it better. How would the number (5 followed by 163 zeros) be expressed as 10^y and how would this be calculated. (using a calculator) Maybe you did understand the question and I just didn't understand the answer. If that's the case, I apologize.

Thanks.

 

[HallsofIvy]

I am inclined to think that your calculator actually read "5.00000000 E+163" and you missed seeing the decimal point.

 

[oupavios]

I am inclined to think that your calculator actually read "5.00000000 E+163" and you missed seeing the decimal point.

You are correct.

 

[Ishuda]

I had another question which should be simple for you guys. I have a value of 2 ^149 which I know from a book I'm reading is roughly 10 ^45. But what calculation needs to be done in order to arrive at the second number?

Thanks again!

As corrected: 2¹⁴⁹ ~ 10⁴⁵

2¹⁴⁹ = 10^x

\(\displaystyle log_{10} 2^{149} = 149 log_{10} 2 = log_{10} 10^x = x \)
or
x ~ \(\displaystyle 149\space log_{10} 2\) ~ 149 * .301 ~ 45
so
2¹⁴⁹ ~ 10⁴⁵