marcelie
  • marcelie
help please !!!! i attached the picture below
Mathematics
schrodinger
  • schrodinger
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marcelie
  • marcelie
|dw:1444196703724:dw|
marcelie
  • marcelie
|dw:1444196743208:dw|
jojokiw3
  • jojokiw3
ooo, so logs are like exponents, when you multiply exponents, what do you do with them actually?

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marcelie
  • marcelie
yea
jojokiw3
  • jojokiw3
What's the answer to this question? x^2(x^2)?
marcelie
  • marcelie
the answer in my sheet says x cant equal -3 and x = 2
jojokiw3
  • jojokiw3
Hm, I don't think you can get an numerical answer to this question, because it doesn't have an equal sign.
jojokiw3
  • jojokiw3
But you can simplify the equation more
marcelie
  • marcelie
yeh one sec ..
freckles
  • freckles
|dw:1444197015545:dw|
marcelie
  • marcelie
1 Attachment
freckles
  • freckles
\[\log_2(x+2)+\log_2(x-1)=2?\]
marcelie
  • marcelie
yes.
jojokiw3
  • jojokiw3
ooo, that makes more sense.
freckles
  • freckles
that first step you have there looks great little confused on your third line you have you should do 2^ on both sides for your second line
marcelie
  • marcelie
the other one should be log base 2 (2) ?
marcelie
  • marcelie
k
freckles
  • freckles
\[2^{\log_2((x-1)(x+2))}=2^2\] see we do 2^ on both sides
freckles
  • freckles
this gives us: \[(x-1)(x+2)=4\]
jojokiw3
  • jojokiw3
Oh, I'm sorry.
freckles
  • freckles
you will need to expand and get everything on one side like you are going to want 0 on that right hand side there
marcelie
  • marcelie
hmmm. im confused
freckles
  • freckles
on what I did?
freckles
  • freckles
or what I'm asking you to do?
marcelie
  • marcelie
yes. why 2 on both sides
freckles
  • freckles
\[a^{\log_a(y)}=y \\ \text{ for } y>0 \text{ and } a \in (0,1) \cup (1,\infty)\]
freckles
  • freckles
\[\text{ that is I knew } 2^{\log_2((x+2)(x-1))}=(x+2)(x-1)\]
freckles
  • freckles
for example if we had: \[\log_{4}((x+2)(x-1))=2 \\ \text{ I would do 4^ on both sides } \\ 4^{(\log_4(x+2)(x-1))}=4^2 \\ \text{ which the whole point in doing the 4^ } \\ \text{ is written below } \\ (x+2)(x-1)=4^2 \\ \text{ we get \to write in its equivalent exponent form } \]
freckles
  • freckles
do you understand that the following two are inverses: \[y=a^x \text{ and } y=\log_a(x)\]
freckles
  • freckles
\[f(x)=a^x \text{ and } f^{-1}(x)=\log_a(x) \text{ are inverses iff } \\ f(f^{-1}(x))=x \text{ and } f^{-1}(f(x))=x\]
freckles
  • freckles
that is: \[a^{\log_a(x)}=x \text{ and } \log_a(a^x)=x\] there are some domain restrictions but yeah
freckles
  • freckles
\[\text{ so we had } \log_2(stuff)=2 \\ \text{ and I did } \color{red}{2}^{\log_\color{green}{2}(stuff)}=\color{red}{2}^2 \\ stuff=4\]
freckles
  • freckles
I did 2^ on both sides because of the green 2 there
freckles
  • freckles
if the green would have been 5 I would have did 5^ on both sides
marcelie
  • marcelie
woahhh o.o
freckles
  • freckles
if that doesn't make sense to you I'm sure you have written log equations in their equivalent exponential form right?
freckles
  • freckles
\[\log_a(y)=x \implies a^{x}=y \\ \log_2([(x-1)(x+2)]=2 \implies 2^{2}=(x-1)(x+2)\]
marcelie
  • marcelie
oh got it xD
marcelie
  • marcelie
thank you :)
freckles
  • freckles
np

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