mathmath333
  • mathmath333
Frustrating problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\normalsize \color{black}{\begin{align}& \text{if}\ \log 242=a,\log 80=b\ \text{and}\ \log 45=c & \hspace{.33em}\\~\\ & \text{express}\ \log36\ \text{in terms of }\ a,b,\ \text{and }\ c.\hspace{.33em}\\~\\ & a.)\ \dfrac{(c-1)(3c+b-4)}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{(c-1)(3c+b-4)}{3} \hspace{.33em}\\~\\ & c.)\ \dfrac{(c-1)(3c-b-4)}{2} \hspace{.33em}\\~\\ & (d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
phi
  • phi
one thought is to first find the prime factors of each of the given numbers
mathmath333
  • mathmath333
i tried and ended up blundering every time in calculation

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mathmath333
  • mathmath333
further i need to solve these problem within 2-3 min
phi
  • phi
are you allowed a calculator?
mathmath333
  • mathmath333
lol no / cheating prohibited
imqwerty
  • imqwerty
try out the options lol
mathmath333
  • mathmath333
of cos
imqwerty
  • imqwerty
so the answer is.......
mathmath333
  • mathmath333
thats a liitle time tough
mathmath333
  • mathmath333
lol and the book gives none of these at last when i checked
imqwerty
  • imqwerty
lol xD i hate when it happens xD
mathmath333
  • mathmath333
the objective here is to i need to observe this problem deeply
imqwerty
  • imqwerty
nd i m the guy who likes to do the opposite of what is said to be done. XD
mathmath333
  • mathmath333
what is the minimum time this problem should be solved
imqwerty
  • imqwerty
1-1.5min
mathmath333
  • mathmath333
i guess there should be some trick
imqwerty
  • imqwerty
there is :)
ChillOut
  • ChillOut
Is the anwser d?
Lurker
  • Lurker
|dw:1438120503037:dw|
ChillOut
  • ChillOut
Please note that: 242= 11²*2, 80=8*10 and 45=90/2
phi
  • 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)
mathmath333
  • mathmath333
i guess this is the only way
ChillOut
  • ChillOut
I can't find a fast way either.
phi
  • 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
ChillOut
  • ChillOut
b=3log2+log10 = 3log2 +1 b-1=3log2
ChillOut
  • ChillOut
c=log45 -> log (90/2)=log90 - log2 c=2log 3+1-log2 c-1=2log3-log2
ChillOut
  • ChillOut
b-1=3log2 (b-1)/3=log2 c-1=2log3-(b-1)/3
imqwerty
  • imqwerty
1 Attachment
ChillOut
  • ChillOut
That's it! This is all about getting the smallest common divisors in terms of log(10)
imqwerty
  • imqwerty
but still i think that tryin out options will be the best
phi
  • phi
to do that in 2 minutes you will have to get *very* familiar with logs
Loser66
  • Loser66
To me, it is "none of them"
mathmath333
  • 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
phi
  • phi
yes , none of the above. But how do you know that quickly?
Loser66
  • Loser66
log (80) = log (20*4) = log (20) +log(4) hence log (4) = log(80) - log (20) = b- log(20)
Loser66
  • 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.

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