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Its a log problem which has sqrt so i shall post in reply...

Mathematics
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\[\log_{4} \sqrt[4]{32\sqrt[3]{16\sqrt{64.8^{-3\div4}}}}\]
2^5 = 32 2^4 = 16 Is it 64 x 8? or 64.8
its 64 x 8^-3/4

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Other answers:

please give full explanation
2^3 = 8 2^6 = 64 Now try it
I don't like this problem
wait even i dont like it
It's not hard but it's unnecessary
its necessary for me
1.019
i am not getting it can u xplain
oh cool sheg
no sheg answer is wrong
how come plz explain m right
idk answer given as 121/186 or 0.6505376
ok wait i made one mistake now i got it
I am getting .7 after simplifying and calculating it in the calculator
do not use claculator
and explain
u should get it as 121/186
King It's a big waste of time, it's not worth it
pls ishaan its necessary fr me
sheg wat abt u?
trying to solve it manually
Hmm I am not posting the complete solution, Just a little simplification \[\frac{1}{4} \left(\log_4 32 + \frac{1}{3}\left(\log_416 + \frac{1}{2}\left(\log_4 64 + \log_4 8^{\frac{-3}{4}} \right)\right)\right)\]
is that 64 times 8?????????????
Now you should do it on your own, I really don't like calculating Numbers
how did u get 1/4
yeah sheg
oh got i was taking 64.8
Yeah Sheg
Read the FIITJEE Module for Logs it's the basic property of Log
nahi yar mein usko decimal mein le raha tha na bracket use kiya hai isne na kuch
wait im reading and i gave u this sum from FIITJEE compendium
sry sheg are u now getting right answer?
Hmm mujhe bhi same confusion hua tha
i m getting this thing \[\log_{4} 2^{\frac{167}{96}}\]
what is that 2^?
167/96
thats 0.8 sumthing...
ishaan i calculated ure equation i got 29/32
you must have calculated it wrong
wait i am checking for mistake
I am getting 127/192 I hate this problem, I hate calculating numbers
.661458333 <
u need to subtract 6 frm numerator and denom. to get rite answer so thatt means u are very close
Took me forever to get to the bottom of this >.>
so post solution katrina
ishaan what is simplified version of log8^-3/4 base4
\[\log_4\sqrt[4]{32\sqrt[3]{16\sqrt{64\cdot8^{-\frac{3}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^6\cdot2^{-\frac{9}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^{\frac{15}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\cdot{2^{\frac{15}{8}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{{2^{\frac{47}{8}}}}}=\] \[\log_4\sqrt[4]{2^5\cdot{{2^{\frac{47}{24}}}}}=\] \[\log_4\sqrt[4]{{{2^{\frac{167}{24}}}}}=\] \[\log_{2^2}2^{\frac{167}{96}}=\] \[\log_{2}2^{\frac{167}{192}}=\] \[\frac{167}{192}\]
Yeah I did it again I am getting the same
can u explain the 4th step
how did u take out 2^15/4?
\[2^4\cdot2^{\frac{15}{8}}=2^{4+\frac{15}{8}}=2^{\frac{32}{8}+\frac{15}{8}}=2^{\frac{47}{8}}\] this one?
no how did u take 2^15/4 out frm sqrt
http://www.wolframalpha.com/input/?i=log_4%28%2832+*+%2816+*%2864+*+8^%28-3%2F4%29%29^%281%2F2%29%29^%281%2F3%29%29^%281%2F4%29 Tomas Is Right, FIITJEE is wrong!
ok now can u xplain
\[\sqrt{2^\frac{15}{4}}=2^{\frac{15}{4}\cdot\frac{1}{2}}=2^\frac{15}{8}\]
oh right
\[\sqrt[n]{m}=m^{\frac{1}{n}}\]
Yipppeeeeeeeee u guys are great!!!!!!!!!! thnx to all of u.!!!!!!!!!!!!!!!

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