Mathematics
6 Online
OpenStudy (anonymous):
solve for x. Round your answer to 2 decimal places.
11-2*log(x)=14
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OpenStudy (anonymous):
I know you have to subtract 11 from both sides 1st
OpenStudy (anonymous):
so it'd be:
-2*log(x)=11
OpenStudy (anonymous):
so then do I divide both sides by -2?
OpenStudy (anonymous):
so it would be log(x)= -5.5???
OpenStudy (blurbendy):
when you subtract 11, you should get
2^log(x) = 3
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OpenStudy (blurbendy):
2*log(x) = 3, dont know where ^ came from
OpenStudy (anonymous):
why would it be 2^??
OpenStudy (anonymous):
oh ok :)
OpenStudy (blurbendy):
also there should be a - in front of the 2
OpenStudy (blurbendy):
so now you can divide both sides by -2
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OpenStudy (anonymous):
sorry that's what I meant
OpenStudy (anonymous):
3/-2 not 11/ -2
OpenStudy (anonymous):
so it'd be log(x)= -1.5
OpenStudy (blurbendy):
yes. no, do you remember how to cancel logs, so you're just left with x?
OpenStudy (blurbendy):
now*
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OpenStudy (anonymous):
umm... so........
OpenStudy (anonymous):
no :(
OpenStudy (blurbendy):
if you take both sides to the e power you get:
e^(log(x) = e^(-1.5)
e^(log) just = 1, so you have
x = e^(-1.5)
OpenStudy (anonymous):
so 31.62?
OpenStudy (anonymous):
wait no
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OpenStudy (blurbendy):
e^(-1.5) = e^(-3/2) = 1 / e^(-3/2)
OpenStudy (blurbendy):
gonna be a small number
OpenStudy (anonymous):
4.48?
OpenStudy (anonymous):
but that would just be e^ positive 1.5 right?
OpenStudy (blurbendy):
sorry, 1 / e^(3/2)
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OpenStudy (blurbendy):
when you move it down to the denominator, the negative sign goes away.
OpenStudy (anonymous):
right because u multiply by the reciprocal?
OpenStudy (anonymous):
so it is 4.48 right?
OpenStudy (blurbendy):
no, 1/e^(3/2) is barely over 0
try again
OpenStudy (anonymous):
.22 ?
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OpenStudy (blurbendy):
yes!
OpenStudy (anonymous):
ok i'm gonna submit it
OpenStudy (anonymous):
& what abt:
e^(0.5x) = 0.5
OpenStudy (blurbendy):
to get rid of log, we took it to the e power. how do you think we get rid of e?
OpenStudy (anonymous):
uh...
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OpenStudy (anonymous):
root it?
OpenStudy (blurbendy):
no, just take the log of both sides:
log(e^(0.5x)) = log(0.5)
log (e) = 1 so you have
0.5x = log(0.5)
OpenStudy (anonymous):
so the log(.5)/ .5 ???
OpenStudy (blurbendy):
mmhm
OpenStudy (anonymous):
I got -0.60
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OpenStudy (anonymous):
but then idk where I got -1.39 earlier.... :(
OpenStudy (blurbendy):
-1.39 is right!
OpenStudy (anonymous):
oh... ok then