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log-5x=log (10-3x)

Mathematics
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\[\log (-5x) = \log(10 - 3x)\] Is that the question?
yeah haha my bad
Do you agree that -5x = 10 - 3x ?

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Other answers:

x=-5?
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
yeah -5 is what i got (:
I don't think you can have a negative log tho :/ so no real solutions?
but RHS is imaginary!
true
so,no real solution exists.
log a = log(4a - 9) Now: a = 4a - 9 -3a = -9 a = 3
OH dude. i'm dumb. for some reason i was moving it over and changing it to -4a -.- thank you so much !!!!
you are not dumb,silly mistakes happens :-)
(: thanks !
haha, the mistakes I've made in math are far worse :) Hope this helped!
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?
\[-4\log(3-9m)=-4 \] ?
\[-4\log _{3}-9m=-4\] (:
wait, that doesn't make sense. \[-4\log_3(-9m)=-4 \]
maybe that's how it's supposed to be. it just doesn't show up like that on the worksheet.
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: \[\log_3(-9m) = 1 \rightarrow 3^1 = -9m\] \[m={-1 \over 3}\]
WOW. you're great.
:) Know the laws by heart, and it will be very very easy!
haha i hope so.

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