How to calculate 167^0,2 without a calculator

  • Thread starter Thread starter kev.thomson96
  • Start date Start date
  • Tags Tags
    Calculator
AI Thread Summary
To calculate 167^0.2 without a calculator, one can express 0.2 as 1/5, leading to the fifth root of 167. The method involves estimating the fifth root through trial values, starting with 2.5 and adjusting upwards until reaching sufficient accuracy. Alternatively, logarithms can be used by finding log(167), multiplying by 0.2, and then determining the antilogarithm, though this approach may require additional steps for numbers outside standard log tables. The discussion also touches on the nostalgia for older calculation tools like slide rules and log tables. Overall, both estimation and logarithmic methods provide viable paths for manual calculation.
kev.thomson96
Messages
13
Reaction score
0
I was wondering how that could be done.


I tried 0,2 ---> 1/5, then 167 ^(1/5) , which should lead to 5√167 (fifth root of 167) , but I can't seem to move on from there
 
Mathematics news on Phys.org
0,2= 1/5 so you are asking for the principal fifth root of 167. I note that 2^5= 32 and 3^5= 243 so I would next try 2.5. 2.5^5= 97.65625 (yes, I did that "by hand"!). That's less than 167 so I would try 2.75 next and keep going until I got sufficient accuracy.
 
I did it using the first two terms of a binomial expansion.

$$167^\frac{1}{5} = \left( 243 - 76 \right) = 243^\frac{1}{5} \left(1 - \frac{76}{243} \right)^\frac{1}{5} \doteq 3 -\frac{1}{5} \frac{76}{81}$$
 
Kos Drago said:
I was wondering how that could be done.


I tried 0,2 ---> 1/5, then 167 ^(1/5) , which should lead to 5√167 (fifth root of 167) , but I can't seem to move on from there

Are you allowed to use log tables?
 
I don't think so, but I'd like you to elaborate if you can solve it with log.
 
If x= 167^{0,2} the log(x)= 0,2 log(167).

So: look up the logarithm of 167 in your log table, multiply by 0,2 then look up the number whose logarithm is that.
 
Note: you will not find logarithm of just 167. log table I have here (base 10) contains logs of numbers between 1 and 10, so you will need to express 167 as 1.67*100 and then log(167) = log(1.67)+2.

Not that it changes the general idea, just makes it a little bit more convoluted.
 
"log table I have here..."

Not sure I could put a finger on a log table if pressed. It reminds me of a conversation I had many years ago (1990-ish) with a historian
Historian: "Do you have a slide rule I can use?"
Me: "No, I haven't had one for many years."
Historian: "I thought every mathematician had one."
Me: "Before you go, do you have any papyrus I could have?"
Historian: "Why would you think we still use that?"

He didn't get my humor.
 
statdad said:
"log table I have here..."

Not sure I could put a finger on a log table if pressed. It reminds me of a conversation I had many years ago (1990-ish) with a historian
Historian: "Do you have a slide rule I can use?"
Me: "No, I haven't had one for many years."
Historian: "I thought every mathematician had one."
Me: "Before you go, do you have any papyrus I could have?"
Historian: "Why would you think we still use that?"

He didn't get my humor.
I haven't used any of mine for some time, but I still have a few slide rules around.
 
  • #10
I still have two, but they are in my house, on the same shelves as the old roll film cameras my father had 85 years ago.
 
Back
Top