## mathmath333 one year ago Frustrating problem

1. mathmath333

\normalsize \color{black}{\begin{align}& \text{if}\ \log 242=a,\log 80=b\ \text{and}\ \log 45=c & \hspace{.33em}\\~\\ & \text{express}\ \log36\ \text{in terms of }\ a,b,\ \text{and }\ c.\hspace{.33em}\\~\\ & a.)\ \dfrac{(c-1)(3c+b-4)}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{(c-1)(3c+b-4)}{3} \hspace{.33em}\\~\\ & c.)\ \dfrac{(c-1)(3c-b-4)}{2} \hspace{.33em}\\~\\ & (d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}

2. phi

one thought is to first find the prime factors of each of the given numbers

3. mathmath333

i tried and ended up blundering every time in calculation

4. mathmath333

further i need to solve these problem within 2-3 min

5. phi

are you allowed a calculator?

6. mathmath333

lol no / cheating prohibited

7. imqwerty

try out the options lol

8. mathmath333

of cos

9. imqwerty

10. mathmath333

thats a liitle time tough

11. mathmath333

lol and the book gives none of these at last when i checked

12. imqwerty

lol xD i hate when it happens xD

13. mathmath333

the objective here is to i need to observe this problem deeply

14. imqwerty

nd i m the guy who likes to do the opposite of what is said to be done. XD

15. mathmath333

what is the minimum time this problem should be solved

16. imqwerty

1-1.5min

17. mathmath333

i guess there should be some trick

18. imqwerty

there is :)

19. ChillOut

Is the anwser d?

20. anonymous

|dw:1438120503037:dw|

21. ChillOut

Please note that: 242= 11²*2, 80=8*10 and 45=90/2

22. phi

I don't see a fast way. but we can look at the first choice and say log(45/10) ( log(45^3) + log(80)-log(10000)) = log(36^2) or log(45/10) ( log ( 5^3 * 3^6 * 2^3 / (2^2 5^2) ) = ... log(4.5) log(2*5*3^6) = log(2^4 3^4)

23. mathmath333

i guess this is the only way

24. ChillOut

I can't find a fast way either.

25. phi

log (9/2) is about 0.95-0.3 = 0.65 or about 2/3 so log(2^(2/3) 5^(2/3) * 3^4) = log(2^4 * 3^4) log(10^(2/3) ) + log(3^4) = log(2^4) + log(3^4) 2/3 + log(3^4) = log(2^4) + log(3^4) 2/3 = log(2^4) ? definitely not

26. ChillOut

b=3log2+log10 = 3log2 +1 b-1=3log2

27. ChillOut

c=log45 -> log (90/2)=log90 - log2 c=2log 3+1-log2 c-1=2log3-log2

28. ChillOut

b-1=3log2 (b-1)/3=log2 c-1=2log3-(b-1)/3

29. imqwerty

30. ChillOut

That's it! This is all about getting the smallest common divisors in terms of log(10)

31. imqwerty

but still i think that tryin out options will be the best

32. phi

to do that in 2 minutes you will have to get *very* familiar with logs

33. Loser66

To me, it is "none of them"

34. mathmath333

i nned to remember log 2 and log 3 values and then try out with options simpifying it only log 2 and log 3

35. phi

yes , none of the above. But how do you know that quickly?

36. Loser66

log (80) = log (20*4) = log (20) +log(4) hence log (4) = log(80) - log (20) = b- log(20)

37. Loser66

36 = 4*9 log 36 = log 4 + log9 = b-log 20 + log9 and log 45 = log 9+ log 5--> log 9 = log 45-log5 = c - log5 hence log36 = b-log20+c-log5 log 36 = b+c-(log20+log5) = b+c-log(20*5) = b+c - log 100 = b+c-2 None of the above can simplify to get b+c-2.