Think you can do logs? | Scoins.net | DJS

## Think you can do logs?

In the Y8 extension work there is an exploration of a^b^c. This is defined as a^(b^c).

Q1. Show that the largest 3ˣ on your calculator is 3^209. Use logs to explain why that is.

Q2. Find the largest 4ˣ your calculator will provide.

Q3. Find the biggest integer n such that 6ⁿ < 12^144   (the biggest two-digit number in duodecimal is ee, 143 base 10). Can you write this 6ⁿ in base 12?   {....,8,9,t,e,10,11,....}

Q4.  If index to two digits worked in hexadecimal on your calculator, what would be the biggest 5^n the calculator would display? (FF<100 in hex). Attempt to write this result in hexadecimal, too.

Q5. Write the whole of 3^3^3. You will need some non-calculus work to find the last two digits.

Q6. By taking logs twice, find the integer N for which N^N^N most closely matches a googolplex, 10^10^100. To justify your chosen answer, find the error, expressed in standard form.  From Large Numbers in the 'Sixth form extension work'. Find the next significant digit and so give N to 3 s.f.

A1. 3^209 < 10^100; 3^210>10^100.   100 / (log3) = 209.59. Calcvulator won't show a third index digit.

A2. 100/log 4 = 166.096 so 4^166 succeeds, approximately 8.75x10^99

A3. 6^n<12^144 so n<144 log12/log6 = 199.7

6^199 = 3.38205 x 10^143 which is, I think 6^147₁₂   = 2.6183₁₂ x 12^ee in base 12

A4.  5^n < 16^100   n log5 < FF log16 = 255₁₀ log 16 => n< 1255 log16/log 5 = 439.29. So the liimiting value is 5^439 which is perhaps 2.07Cx16^FE (that last bit is quite hard and I'm not at all certain`I'm right, so it's not part of the question.)

A5.   7 625 597 494 987  The last two digits are 87, explained on the y8 page.

A6.  51 log (51 log51) = 98.937;  52 log (52 log 52) = 101.427.

51^51^51 = 10^10^98.937 so error is 10^10^1.063=  10^11.56 = 3.6x10^11

52^52^52 has error 10^10^1.427 = 5.37 x 10^26. So 51 is the closer answer

51.4273 gives 100.00001985. the 3 s.f. answer is 51.4.

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