King
Its a log problem which has sqrt so i shall post in reply...
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King
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\[\log_{4} \sqrt[4]{32\sqrt[3]{16\sqrt{64.8^{-3\div4}}}}\]
Ishaan94
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2^5 = 32
2^4 = 16
Is it 64 x 8? or 64.8
King
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its 64 x 8^-3/4
King
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please give full explanation
Ishaan94
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2^3 = 8
2^6 = 64
Now try it
Ishaan94
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I don't like this problem
King
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wait even i dont like it
Ishaan94
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It's not hard but it's unnecessary
King
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its necessary for me
sheg
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1.019
King
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i am not getting it can u xplain
Ishaan94
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oh cool sheg
King
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no sheg answer is wrong
sheg
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how come plz explain m right
King
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idk answer given as 121/186 or 0.6505376
sheg
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ok wait i made one mistake now i got it
Ishaan94
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I am getting .7 after simplifying and calculating it in the calculator
King
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do not use claculator
King
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and explain
King
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u should get it as 121/186
Ishaan94
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King It's a big waste of time, it's not worth it
King
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pls ishaan its necessary fr me
King
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sheg wat abt u?
sheg
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trying to solve it manually
Ishaan94
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Hmm I am not posting the complete solution, Just a little simplification
\[\frac{1}{4} \left(\log_4 32 + \frac{1}{3}\left(\log_416 + \frac{1}{2}\left(\log_4 64 + \log_4 8^{\frac{-3}{4}} \right)\right)\right)\]
sheg
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is that 64 times 8?????????????
Ishaan94
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Now you should do it on your own, I really don't like calculating Numbers
King
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how did u get 1/4
King
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yeah sheg
sheg
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oh got i was taking 64.8
Ishaan94
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Yeah Sheg
Ishaan94
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Read the FIITJEE Module for Logs it's the basic property of Log
sheg
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nahi yar mein usko decimal mein le raha tha na bracket use kiya hai isne na kuch
King
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wait im reading and i gave u this sum from FIITJEE compendium
King
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sry sheg are u now getting right answer?
Ishaan94
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Hmm mujhe bhi same confusion hua tha
sheg
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i m getting this thing
\[\log_{4} 2^{\frac{167}{96}}\]
King
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what is that 2^?
sheg
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167/96
King
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thats 0.8 sumthing...
King
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ishaan i calculated ure equation i got 29/32
Ishaan94
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you must have calculated it wrong
King
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wait i am checking for mistake
Ishaan94
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I am getting 127/192 I hate this problem, I hate calculating numbers
Ishaan94
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.661458333 <<Calculator
King
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u need to subtract 6 frm numerator and denom. to get rite answer so thatt means u are very close
KatrinaKaif
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Took me forever to get to the bottom of this >.>
King
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so post solution katrina
King
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ishaan what is simplified version of log8^-3/4 base4
Tomas.A
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\[\log_4\sqrt[4]{32\sqrt[3]{16\sqrt{64\cdot8^{-\frac{3}{4}}}}}=\]
\[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^6\cdot2^{-\frac{9}{4}}}}}=\]
\[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^{\frac{15}{4}}}}}=\]
\[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\cdot{2^{\frac{15}{8}}}}}=\]
\[\log_4\sqrt[4]{2^5\sqrt[3]{{2^{\frac{47}{8}}}}}=\]
\[\log_4\sqrt[4]{2^5\cdot{{2^{\frac{47}{24}}}}}=\]
\[\log_4\sqrt[4]{{{2^{\frac{167}{24}}}}}=\]
\[\log_{2^2}2^{\frac{167}{96}}=\]
\[\log_{2}2^{\frac{167}{192}}=\]
\[\frac{167}{192}\]
Ishaan94
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Yeah I did it again I am getting the same
King
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can u explain the 4th step
King
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how did u take out 2^15/4?
Tomas.A
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\[2^4\cdot2^{\frac{15}{8}}=2^{4+\frac{15}{8}}=2^{\frac{32}{8}+\frac{15}{8}}=2^{\frac{47}{8}}\]
this one?
King
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no how did u take 2^15/4 out frm sqrt
King
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ok now can u xplain
Tomas.A
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\[\sqrt{2^\frac{15}{4}}=2^{\frac{15}{4}\cdot\frac{1}{2}}=2^\frac{15}{8}\]
King
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oh right
Tomas.A
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\[\sqrt[n]{m}=m^{\frac{1}{n}}\]
King
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Yipppeeeeeeeee u guys are great!!!!!!!!!!
thnx to all of u.!!!!!!!!!!!!!!!