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Mathematics 6 Online
OpenStudy (anonymous):

solve for x. Round your answer to 2 decimal places. 11-2*log(x)=14

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

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

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*

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

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)

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 ?

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...

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

OpenStudy (anonymous):

but then idk where I got -1.39 earlier.... :(

OpenStudy (blurbendy):

-1.39 is right!

OpenStudy (anonymous):

oh... ok then

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