Question

if log(subscript a) x=7, what is the value of log(subscript a) (a/x^2) ?

Answer 1

\[\Large\bf\sf \log_ax=7,\qquad\qquad \log_a\frac{a}{x^2}=?\]

Answer 2

We can apply some rules of logs to help us out!

Answer 3

I'm all ears

Answer 4

Here is an important rule to remember:\[\Large \bf\sf \color{orangered}{\log\left(\frac{b}{c}\right)\quad=\quad \log (b)-\log (c)}\]

Answer 5

So that will allow us to take the first step in rewriting our expression.\[\Large\bf\sf \log_a\frac{a}{x^2}\quad=\quad \log_aa-\log_ax^2\]

Answer 6

Understand how I applied the rule there? :o

Answer 7

I think so :) ... I'm taking notes.

Answer 8

So where do I go from here?

Answer 9

Whenever the `base` of a log matches the `contents` of the log, the result is 1.
If you don't understand why, then it might seem a little weird.

You can punch it into your calculator to check.
Like if you try \(\Large\bf\sf log_{10}10\) you should get 1.
So we can simplify the first term down.

Answer 10

\[\Large\bf\sf \log_aa-\log_ax^2\quad=\quad 1-\log_ax^2\]

Answer 11

oh ok

Answer 12

Another important log rule:\[\Large\bf\sf \color{royalblue}{\log(b^c)\quad=\quad c \log(b)}\]Allows us to bring an exponent down in front of the log as a coefficient.

Answer 13

Understand how we can apply this to the 2 on x^2?

Answer 14

Yes 2 log x

Answer 15

Ok good!
\[\Large\bf\sf 1-2\log_ax\]Lemme color it a sec, make something will jump out at you.\[\Large\bf\sf 1-2\color{#DD4747 }{\log_ax}\]That is our expression so far.
Here is the initial information they gave us:\[\Large\bf\sf \color{#DD4747}{\log_ax=7}\]

Answer 16

the coloring really helps!

Answer 17

Do you see where we can plug in our 7?

Answer 18

yes so 1-2(7)

Answer 19

Yesss, good good good.

Answer 20

-13?

Answer 21

Yay good job! \c:/

Answer 22

Thanks:) Wait so that's the final answer?

Answer 23

This is a pretty tough problem.
You have to use 3 different log rules, then use the initial data.

It's a bit of a thinker!

Answer 24

Ya that should be right :o

Answer 25

Jeez

Answer 26

Ok because I thought I had to plug it into log a (a/x^2)

Answer 27

and I didn't square anything...

Answer 28

They gave a nice simple number for log_a(x).

So the idea was: we wanted to twist things around to try and get a log_a(x) out of that messy thing :o

Answer 29

ok, so I don't have to use the second part of the problem? sorry I'm just confused :)

Answer 30

Second part of the problem?
You mean this?\[\Large\bf\sf \log_a\frac{a}{x^2}=?\]

Answer 31

yes

Answer 32

We did use that silly!
That's what we `started` with.
And widdled it down doing some steps:\[\Large\bf\sf \log_a\frac{a}{x^2}\quad=\quad\log_aa-\log_ax^2\quad=\quad1-2\log_ax\quad=\quad1-2(7)\]

Answer 33

Oh wow, yes we did.... my bad :)

Answer 34

Can you help me with one more problem?

Answer 35

Mm I can try -_- gotta hit the sack soon. So late!

Answer 36

Explain the algebraic process of how functions can be shown to be inverses. Use the functions below to verify your explanation. f(x) = 3x-1/x+2 and g(x) = 2x+1/3-x

Answer 37

Ok I REALLY appreciate all your help! You are a life saver!!! And I know I'm tired too!

Answer 38

\[\Large\bf\sf f(x)=\frac{3x-1}{x+2},\qquad\qquad g(x)=\frac{2x+1}{3-x}\]Let's fiddle around with f(x) and try to show that g(x) is it's inverse.
We'll start by letting f(x) = y.\[\Large\bf\sf y=\frac{3x-1}{x+2}\]

Answer 39

ok

Answer 40

To find the inverse of this function, we `swap/trade` the x's and y's in the equation.\[\Large\bf\sf x=\frac{3y-1}{y+2}\]

Answer 41

To find the inverse function, we now need to solve for y.

Answer 42

ok

Answer 43

It'll take a few tricky algebra steps to get our y.

Answer 44

We'll multiply both sides by (y+2)

Answer 45

\[\Large\bf\sf x(y+2)=3y-1\]

Answer 46

ok, with u so far

Answer 47

We'll distribute the x to each term in the brackets.

Answer 48

xy + 2x

Answer 49

Ok good.\[\Large\bf\sf xy+2x=3y-1\]

Answer 50

Now we want to get both of the `y` terms on the same side.
So let's subtract 3y from each side.

Answer 51

ok

Answer 52

No no no, let's not do that.
Let's subtract xy from each side actually...
It'll work out a little nicer.

Answer 53

-3y + xy + 2x= -1?

Answer 54

ok

Answer 55

so what does our equation look like now?

Answer 56

So I guess that gives us this for now:
\[\Large\bf\sf 2x=3y-xy-1\]

Answer 57

ok

Answer 58

We'll add 1 to each side,\[\Large\bf\sf 2x+1=3y-xy\]Then we'll want to `factor out` a y from each term on the right.

Answer 59

ok

Answer 60

\[\Large\bf\sf 2x+1=y(3-x)\]Understand the factoring ok?

Answer 61

pooh yes, I was doing in wrong on my paper, but yes i see

Answer 62

*oooh

Answer 63

lol XD dat pooh bear

Answer 64

:)

Answer 65

So to get y alone, let's divide the junk over to the other side.

Answer 66

ok so 2x+1/y = 3-x?

Answer 67

Woops! We don't want to divide the y to the left side silly!
We're trying to get y alone.
I was referring to the `(3-x)` as the junk connected to y.

Answer 68

ok, so what is our equation now?

Answer 69

2x+1/3-x=y

Answer 70

got it

Answer 71

\[\Large\bf\sf \frac{2x+1}{3-x}=y\]Yes, good good.
Our final step is to rewrite y as \(\Large\bf\sf f^{-1}(x)\).
This new y that we've found now represents the inverse function of f, since we swapped the x's and y's

Answer 72

oooohhhh and this equals the other equation we were given!!! so they are inverses of each other, right?

Answer 73

Yesss, good!
So we've just found out that \(\Large\bf\sf f^{-1}(x)=g(x)\)
Which means f and g are inverse functions of one another! :O

Answer 74

I cannot even began to thank you enough for all your help!!!!!

Answer 75

lol np :3
go to bed bunny it's late!! +_+ err at least in this part of the world it is. :P

Answer 76

No its past midnight here! Thx a million!