help please !!!! i attached the picture below

- marcelie

help please !!!! i attached the picture below

- schrodinger

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- marcelie

|dw:1444196703724:dw|

- marcelie

|dw:1444196743208:dw|

- jojokiw3

ooo, so logs are like exponents, when you multiply exponents, what do you do with them actually?

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## More answers

- marcelie

yea

- jojokiw3

What's the answer to this question? x^2(x^2)?

- marcelie

the answer in my sheet says x cant equal -3 and x = 2

- jojokiw3

Hm, I don't think you can get an numerical answer to this question, because it doesn't have an equal sign.

- jojokiw3

But you can simplify the equation more

- marcelie

yeh one sec ..

- freckles

|dw:1444197015545:dw|

- marcelie

##### 1 Attachment

- freckles

\[\log_2(x+2)+\log_2(x-1)=2?\]

- marcelie

yes.

- jojokiw3

ooo, that makes more sense.

- 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

the other one should be log base 2 (2) ?

- marcelie

k

- freckles

\[2^{\log_2((x-1)(x+2))}=2^2\]
see we do 2^ on both sides

- freckles

this gives us:
\[(x-1)(x+2)=4\]

- jojokiw3

Oh, I'm sorry.

- 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

hmmm. im confused

- freckles

on what I did?

- freckles

or what I'm asking you to do?

- marcelie

yes. why 2 on both sides

- freckles

\[a^{\log_a(y)}=y \\ \text{ for } y>0 \text{ and } a \in (0,1) \cup (1,\infty)\]

- freckles

\[\text{ that is I knew } 2^{\log_2((x+2)(x-1))}=(x+2)(x-1)\]

- 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

do you understand that the following two are inverses:
\[y=a^x \text{ and } y=\log_a(x)\]

- 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

that is:
\[a^{\log_a(x)}=x \text{ and } \log_a(a^x)=x\]
there are some domain restrictions but yeah

- 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

I did 2^ on both sides
because of the green 2 there

- freckles

if the green would have been 5
I would have did 5^ on both sides

- marcelie

woahhh o.o

- freckles

if that doesn't make sense to you
I'm sure you have written log equations in their equivalent exponential form
right?

- freckles

\[\log_a(y)=x \implies a^{x}=y \\ \log_2([(x-1)(x+2)]=2 \implies 2^{2}=(x-1)(x+2)\]

- marcelie

oh got it xD

- marcelie

thank you :)

- freckles

np

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