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how can i do this math shit?
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| Psychonoise |
| 3*4^x/2*9^x=2 |
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| MERTON |
| just figure out what x is.. i haven't gotten to variable exponents yet. |
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| Psychonoise |
| well i tried but my answer was 2^x=2*3^x |
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| MERTON |
| that's not an answer. solve for x! |
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| montie |
well whenever you divide two numbers which are raised to the same exponet, it is equal to the fraction to that exponet
3*(4^x)/2*(9^x)=2 goes to (3/2)*(4/9)^x=2 then (4/9)^x=4/3 then x*log(4/9)=log(4/3) divide the logs x=log(4/3)/log(4/9)
you get some crazy -.3 radical number |
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| crchingtiger |
| i've got it down to 36^x=1/3 |
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| paranoik0 |
| i think that's impossible. for positive numbers at least. try coming up with values and replacing. but probably you need the full resolution of the exercise, right? |
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| montie |
its not impossible. i just solved it -.3547556457 and so forht.
plug and chug and you will see it checks out |
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| MK-S |
i used logs and got x=1.505 (4 s.f)
completely wrong \o/ |
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| Psychonoise |
whole ing !!This is a exercise of school from my 16 Y/O sister!!
i know my english is poor |
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| crchingtiger |
my suggestion was completely wrong! sorry...
hi. my name is laurel and i'm an idiot.
damn math !
it's prolly some weird ass fraction |
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| Psychonoise |
| hehehe ok...math sumetimes suxs... |
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