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- anonymous

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

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- anonymous

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

- jamiebookeater

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- slaaibak

\[\log (-5x) = \log(10 - 3x)\]
Is that the question?

- anonymous

yeah haha my bad

- slaaibak

Do you agree that
-5x = 10 - 3x ?

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- anonymous

x=-5?

- anonymous

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

- anonymous

yeah -5 is what i got (:

- slaaibak

I don't think you can have a negative log tho :/ so no real solutions?

- anonymous

but RHS is imaginary!

- anonymous

true

- anonymous

so,no real solution exists.

- slaaibak

log a = log(4a - 9)
Now:
a = 4a - 9
-3a = -9
a = 3

- anonymous

OH dude. i'm dumb. for some reason i was moving it over and changing it to -4a -.- thank you so much !!!!

- anonymous

you are not dumb,silly mistakes happens :-)

- anonymous

(: thanks !

- slaaibak

haha, the mistakes I've made in math are far worse :) Hope this helped!

- anonymous

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

\[-4\log(3-9m)=-4 \]
?

- anonymous

\[-4\log _{3}-9m=-4\] (:

- slaaibak

wait, that doesn't make sense.
\[-4\log_3(-9m)=-4 \]

- anonymous

maybe that's how it's supposed to be. it just doesn't show up like that on the worksheet.

- slaaibak

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:
\[\log_3(-9m) = 1 \rightarrow 3^1 = -9m\]
\[m={-1 \over 3}\]

- anonymous

WOW. you're great.

- slaaibak

:) Know the laws by heart, and it will be very very easy!

- anonymous

haha i hope so.

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