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OpenStudy (anonymous):
Find the value of k.
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OpenStudy (anonymous):
OpenStudy (anonymous):
The value of k where?
OpenStudy (anonymous):
From the attached picture @mightyrecords
OpenStudy (anonymous):
What i have done is swapped the 3^k with the 42(3^3/2). so i got something like
3^7/2+3^5/2+3^3/2-3^k=42(3^3/2). are we okay upto that point
OpenStudy (anonymous):
maths made simpler. wish i had that application for writing answers with a stylus. pc's suck in maths!
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OpenStudy (anonymous):
ktnguyen, the 42u in the first equations has power 3
OpenStudy (anonymous):
The answer is given as 4.5....
OpenStudy (anonymous):
Please help... :/
OpenStudy (dls):
I got it :)
OpenStudy (anonymous):
please show. are you still using logs and indices?
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OpenStudy (dls):
\[(\sqrt(3))^7+(\sqrt(3))^5+(\sqrt(3))^3+42(\sqrt 3) \]
You can solve this part I guess,you will get around 140.3 u can round it off to 140
OpenStudy (dls):
now 140 is 3 to the power what?
we know
3^2=9
3^3=27
3^4=81
3^5=243
oops,so 140 lies somewhere betwen 4 and 5 so its 4.5
just a layman basic maths :)
OpenStudy (anonymous):
If you use logs from DLS's answer you will get that exact answer 4.5
OpenStudy (anonymous):
it will be something like (log 140.3)/log3=k
OpenStudy (dls):
@BelleFlower ?
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OpenStudy (raden):
|dw:1354956137036:dw|
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