Write the expression below so that only a single logarithm or exponential function appears.
(1/2)ln(z) − ln(5 + x) − 4 ln(y)

- anonymous

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- zepdrix

Rule of logs:\[\Large \color{royalblue}{b\cdot\log(a) \quad=\quad \log(a^b)}\]Understand how we can use this rule to deal with the 1/2 coefficient on the first term?

- anonymous

ln(z)^(1/2)

- zepdrix

The exponent is being applied to the z, not the log. Just be careful the way you write that :)
ln(z^(1/2))

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## More answers

- zepdrix

We can do the same with the 4, yes?

- anonymous

yes

- anonymous

ln(z^(1/2))-ln(5-x)-ln(y^4)

- zepdrix

\[\Large \ln \left(z^{1/2}\right) − \ln(5 + x) − \ln\left(y^4\right)\]Ok cool.

- zepdrix

We'll want to apply another rule of logs from here:\[\Large \color{#CC0033}{\log(a)-\log(b) \quad=\quad \log\left(\frac{a}{b}\right)}\]

- anonymous

I get this ln((z^(1/2))/((5+x)(y^4)))

- zepdrix

\[\Large \ln\left[\frac{z^{1/2}}{(5+x)y^4}\right]\]
With the last two terms in the denominator?
Mmmm yah that looks right.
Good job! :)

- anonymous

yay thank you:)

- anonymous

hey zep could u help me with another problem.. i dont know what im doing wrong

- zepdrix

sure

- anonymous

A very small tumor of initial size M is observed every ten days, and over each ten-day period, it has grown 70% larger.
(a) Write a formula representing the size of the tumor after t days
(t = 0, 10, 20, . . .)

- anonymous

(b) On which day will we finally observe that the tumor is more than 10 times its original size? (Assume that the observations are strictly periodically conducted as specified.)

- anonymous

for part a) I'm getting (1.7)^tM and it says is wrong

- zepdrix

Is the M in the exponent? I can't tell with the way you wrote it :3

- anonymous

no multiplying

- anonymous

(1.7^t)*M

- zepdrix

Mmm lemme see if I've got the right idea here.\[\Large f(t)=Mb^t\]Where M is the initial mass of the tumor.
at \(\Large t=10\) the tumor has grown to a size of 170% or \(\Large 1.7M\)
\[\Large 1.7M=Mb^{10}\]

- zepdrix

\[\Large 1.7=b^{10}\]

- zepdrix

\[\Large b\approx 1.054\]

- zepdrix

Do you have a way to check this? To make sure I'm not doing something really stupid :) lol

- anonymous

well i need a formula for part a

- anonymous

and i get (1.7^t)M
and no i dont have a way to check it :(

- zepdrix

The way I did it, we would end up with,\[\Large b=1.7^{1/10}\]Plugging this into our function gives us,\[\Large f(t)=M\left(1.7^{1/10}\right)^t \qquad\to\qquad f(t)=M\left(1.7\right)^{t/10}\]

- zepdrix

^ Does this way make sense? :o
How did you come up with b=1.7?

- anonymous

yup that worked!

- anonymous

you are amazing!

- zepdrix

yay team \c:/

- anonymous

hey for part b do i just plug 10 for t?

- anonymous

and solve for M?

- zepdrix

On which day, \(\large t\), will we finally observe that the tumor, \(\large f(t)\), is more than 10 times, \(\large \gt 10M\) its original size?

- zepdrix

\[\Large 10M

- zepdrix

I guess we don't really need the inequality.
We can just use an = sign.
Whatever t day we end up with, we will round up to the next day to show the day when the mass is 10times GREATER THAN it's initial size.

- anonymous

ok let me try

- anonymous

t is about 43.3936

- anonymous

so 4 days?

- anonymous

44

- zepdrix

ya that sounds right! :)

- anonymous

says its wrong :(

- zepdrix

Hmm thinkinggg

- zepdrix

(Assume that the observations are strictly `periodically` conducted as specified.)
Only periodically?
So does this mean our t MUST BE a multiple of 10 maybe?
Maybe try t=50 :\
If not we'll have to think about it a lil more.

- anonymous

that worked!!!!

- anonymous

thank you sooo much!

- zepdrix

Oh cool! :) Strange wording on the problem. heh

- zepdrix

\[\Large f(\color{#CC0033}{x})=4-4\color{#CC0033}{x}-\color{#CC0033}{x}^2\]
\[\Large f(\color{#CC0033}{3})=4-4\cdot\color{#CC0033}{3}-\color{#CC0033}{3}^2\]\[\Large f(\color{#CC0033}{3+h})=4-4(\color{#CC0033}{3+h})-(\color{#CC0033}{3+h})^2\]

- zepdrix

The next part is a little tricky.
You need to feel somewhat comfortable with function notation.
We're going to plug in the pieces.\[\Large \frac{f(\color{#CC0033}{3+h})-f(\color{#CC0033}{3})}{h}\]

- anonymous

yea im having trouble with this part

- zepdrix

Plugging it all in is a bit of a pain in the butt. We get something like this,\[\large \frac{4-4(3+h)-(3+h)^2-\left[4-4(3)-3^2\right]}{h}\]

- zepdrix

Notice how I put parentheses around the f(3)?
That's important. The negative needs to be distributed to each term in the brackets.

- anonymous

ok
so now i solve for h

- zepdrix

Here's a little tip to try to keep in mind. Store this in the back of your head somewhere, it might come in handy later.
`If you do all of your simplification correctly, every term that doesn't contain an h will cancel out`.

- zepdrix

Not solve for h, we're just trying to simplify the expression.

- zepdrix

Do a bunch of multiplication, expand out the binomial with the square on it, and simplify! :D

- zepdrix

Remember how to expand this out? \(\Large (3+h)^2\)

- anonymous

omg this is going to be a pain... let me try this lol

- anonymous

i get (-36-3h+h^2)/h

- anonymous

not sure on that

- zepdrix

Remember, EVERYTHING that doesn't have an `h` in the numerator should have cancelled out.
So you shouldn't be left with -36. Hmm

- zepdrix

If we ignore the denominator for a sec,\[\large 4-4(3+h)-(3+h)^2-\left[4-4(3)-3^2\right]\]So we get some stuff,\[\large 4-12-4h-9-6h-h^2-4+12+9\]

- zepdrix

See any mistakes you may have made? :o

- anonymous

-12-4h-6h-h^2?

- anonymous

-12-10h-h^2?

- zepdrix

Where is the -12 coming from? :o
See the cancellations?\[\large \cancel4\cancel{-12}-4h\cancel{-9}-6h-h^2\cancel{-4}\cancel{+12}\cancel{+9}\]

- anonymous

omg im blind

- zepdrix

:3

- anonymous

(-10h-h^2)/h

- anonymous

can i simplify more than that?

- zepdrix

yah you can divide an h out of every term.

- zepdrix

Or you can factor an h out of each term in the numerator, if you feel more comfortable doing that step before the division.

- anonymous

-10-h is the final answer?

- zepdrix

yay good job \c:/

- anonymous

yay thank you!

- anonymous

I have a similar question.. could u set it up for me and let me try to do it?

- zepdrix

sure

- zepdrix

working on b?

- zepdrix

I hate that notation for composition, let's use this instead.\[\Large (g\circ f)(4) \quad=\quad g\left[f(4)\right]\]

- anonymous

shouldnt it be -3?

- zepdrix

Work from the inside out.\[\Large g\left[\color{#F35633}{f(4)}\right]\]
\[\Large \color{#F35633}{f(4)=3}\]\[\Large g\left[\color{#F35633}{f(4)}\right] \quad=\quad g\left[\color{#F35633}{3}\right] \quad=\quad ?\]

- anonymous

1?

- zepdrix

yup!

- anonymous

omg!!! ty

- anonymous

u make it look so easy

- anonymous

could you help me with part e) that looks complicated

- zepdrix

\[\Large h(\color{#3399AA}{x})=\sqrt{\color{#3399AA}{x}^2-f(g(\color{#3399AA}{x}))}\]

- zepdrix

\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(g(\color{#3399AA}{2}))}\]

- zepdrix

Again, work from the inside out,\[\Large g(2)=?\]

- anonymous

5?

- zepdrix

\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(5)}\]

- zepdrix

f(5)=?
:o

- anonymous

f(5)=4

- anonymous

sqrt(4-4)

- anonymous

undefined?

- zepdrix

Mmm sqrt0 = 0
cuz 0*0=0

- zepdrix

Or better said: 0^2=0

- anonymous

it says 0 is not the correct answer

- zepdrix

Hmmm :\

- anonymous

what did we do wrong :(

- zepdrix

Did you try undefined yet? :O
Seems like that shouldn't be the answer, but whatev, it's worth trying.

- anonymous

no undefined is not an answer either :(

- zepdrix

Oh I see, that is not the composition operator in part e, my bad.
That's multiplication I think.

- zepdrix

I'm not really familiar with this notation, does it mean this?\[\Large (f\cdot g)(x) \quad=\quad f(x)\cdot g(x)\]

- zepdrix

Umm let's try it and see what happens.

- zepdrix

\[\Large h(\color{#3399AA}{x})=\sqrt{\color{#3399AA}{x}^2-f(\color{#3399AA}{x})\cdot g(\color{#3399AA}{x})}\]

- anonymous

ok

- zepdrix

\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(\color{#3399AA}{2})\cdot g(\color{#3399AA}{2})}\]

- zepdrix

We get this, yes?\[\Large h(2)=\sqrt{2^2-(-1)\cdot(5)}\]

- anonymous

sqrt(-1)?

- anonymous

or sqrt(9)

- zepdrix

ya sqrt9 i think :L

- anonymous

yay that worked!!!!

- zepdrix

yay team!

- anonymous

:)

- anonymous

hey zep do u have time for one more question

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