a^b^c explored | Scoins.net | DJS

## a^b^c explored

I thought I wrote this page some monthe ago but cannot find it so I'll start all over again.

I suspect there may be confusion over multiple index. 2^2^2 is both 2^4 and 4^2, so however you choose to read it, 16. But 3^3^3 might be 27^3 =19683 or it might be 3^27 = 7.62 x10^12.  My calculator says the smaller one, google says the larger.

Let's go back a bit. We're quite clear, I hope that 7³ means 7x7x7 (=343). So if we multiply 7³ by 7⁵ this could have been writeen as eight 7s to multiply together, 7⁸. So we learn to add index in multiplication. We also learned to multiply indecx across brackets: (10³)², or (10^2)^3 means 10³x10³ which is, from the earlier rule, 10⁶. Therefore we have a pair of rules, to add index when multiplying and the mutliply index across brackets.

So what about index raised to further index, without brackets? The exponent power rule says that the non-bracket version assumes that 'the brackets were up in the air', that a^b^c = a^(b^c). So, if all the indices are positive intyegers, you want the very big result. 3^3^3 is pushing eight million milion.

Check exercise: no two answers are the same.

1.   2^3^2                     2.   3^2^3               3.  2^2^3        4.  3^3^2    5.   2^3^3

So if we accept that 3^3^3 is very large, 7 625 597 484 9xx, what is that last digit?

Powers of three end in a cycle of four, {3,9,7,1},  so 3^27 ends with the same as the other powers of 27mod 4, or 3. So the last digit is as fo 3^3 and 3^7, i.e. a 7. I suggest that the last two digits are 87.

The second last digit  cycles too, and for the (3mod4 set this adds six each time (though you might see this as subtract four). therefore this repeats every five, which (5x4=20) means that the last two digits are the smae for 3^27 as they are for 3^7 (=2187) so 87 is correct. So 3^3^3 = 7 625 597 484 987. Which is more digits than my calculator can provide.

Q6. Find 3^3^4.   Predict the last two digits of this number.

Q7. Have a go at 3^3^5. Explain why your calculator refuses to do this. Find the largest 3^n that your calculator will give a result for.

There is a wangle around this. Of course there is. If you look at yopur calculator and find the 10^x button, it is probable that the other function of this key is labelled 'log'. That is, whatever 10^x does, log is the inverse function. so if 3^N might be 10^100 the linverse function would say that N(log3) = 100, so that suggests that N is 100/log3. `Try that on your calculator and compare this with your answer to Q7.

So to find 3^3^5 = 3^243, we write this as 3^243 = nx10^(lots) and then write the 'log' version, that 243 (log3) = n. My calculator says this is 115.940. so this is 10^115.940, which you might translate as 10^ (115+0.94) = 10^0.94 x 10^115. Which explains how much bigger than 10^100 this is. Now we need to decipher 10^0.94. The same key on the calculator does this (with the Shift or Function key as necessary, or your calculator just might have separate keys for log and 10ˣ). On my calculator that bit before the 10^115 works out as 8.72 to 3 s.f. 8.71896 to 6s.f. I did this by returning to the display taht showed me 115.940....., took off the 115 to have 0.940.... and did 10ˣ to that. Which makes sense, I think.

Let's see if we can formalise this. if 10^(something) = 3 then 10^(log3) = 3 and you calculator is quite happy to tell you log3 = 0.4771...  So 10^(log N) = N, so that you can see log of x  as the inverse function of 10ˣ. the cponvention iis to lose brackets, but say either 'log x' or 'the log of x'. Formally, this is 'the log of x in base ten', or 'log base ten of x'.

Q8.  Find 3^3^6 to 6 s.f.

Q9.   Find 4^4^4 to 6 s.f.

Q10.  Find 3^3^3^2 to 2 s.f.

1.   2^3^2  = 2^9 = 512

2.   3^2^3  =3^8 = 6561

3.  2^2^3  = 2^8 = 256

4.  3^3^2  = 3^9 = 19683

5.   2^3^3 = 2^27 = 134 217 728

6.  3^3^4 = 4.43426488 x 10^38  (that's 10³⁸). The last two digits of 3^81 are the same as for 3^61, 3^41 and 3^01. So the number ends 03.

7. 3^3^5 = 3^243. This is bigger than 10^100; calculators only give two digits of index.

3^209 = 5.23x10⁹⁹. 3^210 exceeds 10^100

8. 36(3^6) = 3^729 so we want 729 log3 = 10^347.821394691 = 10^(0.821..)   x10^347

= 6.62819 x 10^347.

9. 4^4^4 = 4^256. 256 log 4 = 154.12735 =   1.34078 x 10^127

10.  3^3^3^2 = 3^3^9 = 3^19683. 19683 (log 3) = 9391.1776 = 1.505464 x 10^9391 Answer 1.5 x 10^9391.

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