taranicolee
log-5x=log (10-3x)
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slaaibak
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\[\log (-5x) = \log(10 - 3x)\]
Is that the question?
taranicolee
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yeah haha my bad
slaaibak
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Do you agree that
-5x = 10 - 3x ?
FoolForMath
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x=-5?
taranicolee
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i understand that one now though. but another one of the questions is log a=og(4a-9) and the answer is supposed to be 3 but i'm getting 2.25
taranicolee
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yeah -5 is what i got (:
slaaibak
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I don't think you can have a negative log tho :/ so no real solutions?
FoolForMath
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but RHS is imaginary!
taranicolee
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true
FoolForMath
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so,no real solution exists.
slaaibak
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log a = log(4a - 9)
Now:
a = 4a - 9
-3a = -9
a = 3
taranicolee
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OH dude. i'm dumb. for some reason i was moving it over and changing it to -4a -.- thank you so much !!!!
FoolForMath
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you are not dumb,silly mistakes happens :-)
taranicolee
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(: thanks !
slaaibak
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haha, the mistakes I've made in math are far worse :) Hope this helped!
taranicolee
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wait i'm sorry can i ask another question? ( i was absent when this was all being taught) but how would you solve
-4log3-9m=-4?
since there's no log on the right side of the equation?
slaaibak
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\[-4\log(3-9m)=-4 \]
?
taranicolee
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\[-4\log _{3}-9m=-4\] (:
slaaibak
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wait, that doesn't make sense.
\[-4\log_3(-9m)=-4 \]
taranicolee
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maybe that's how it's supposed to be. it just doesn't show up like that on the worksheet.
slaaibak
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Assuming that's the question:
First divide both side by -4:
\[\log_3(-9m) = 1\]
Now, you need to know this property of logarithms:
\[a^x = b\]
\[\log_a b = x\]
\[a^x = b\]
Now your problem:
\[\log_3(-9m) = 1 \rightarrow 3^1 = -9m\]
\[m={-1 \over 3}\]
taranicolee
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WOW. you're great.
slaaibak
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:) Know the laws by heart, and it will be very very easy!
taranicolee
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haha i hope so.