Question

Write the expression below so that only a single logarithm or exponential function appears.

(1/2)ln(z) − ln(5 + x) − 4 ln(y)

Answer 1

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?

Answer 2

ln(z)^(1/2)

Answer 3

The exponent is being applied to the z, not the log. Just be careful the way you write that :)

ln(z^(1/2))

Answer 4

We can do the same with the 4, yes?

Answer 5

yes

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

yay thank you:)

Answer 12

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

Answer 13

sure

Answer 14

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, . . .)

Answer 15

(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.)

Answer 16

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

Answer 17

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

Answer 18

no multiplying

Answer 19

(1.7^t)*M

Answer 20

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 \(\Large 1.7M\)

\[\Large 1.7M=Mb^{10}\]

Answer 21

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

Answer 22

\[\Large b\approx 1.054\]

Answer 23

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

Answer 24

well i need a formula for part a

Answer 25

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

Answer 26

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

Answer 27

^ Does this way make sense? :o

How did you come up with b=1.7?

Answer 28

yup that worked!

Answer 29

you are amazing!

Answer 30

yay team \c:/

Answer 31

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

Answer 32

and solve for M?

Answer 33

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?

Answer 34

\[\Large 10M<M\left(1.7\right)^{t/10}\]

Answer 35

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.

Answer 36

ok let me try

Answer 37

t is about 43.3936

Answer 38

so 4 days?

Answer 39

44

Answer 40

ya that sounds right! :)

Answer 41

says its wrong :(

Answer 42

Hmm thinkinggg

Answer 43

(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.

Answer 44

that worked!!!!

Answer 45

thank you sooo much!

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

yea im having trouble with this part

Answer 50

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

Answer 51

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

Answer 52

ok
so now i solve for h

Answer 53

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`.

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

not sure on that

Answer 60

Remember, EVERYTHING that doesn't have an `h` in the numerator should have cancelled out.

So you shouldn't be left with -36. Hmm

Answer 61

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,\[\large 4-12-4h-9-6h-h^2-4+12+9\]

Answer 62

See any mistakes you may have made? :o

Answer 63

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

Answer 64

-12-10h-h^2?

Answer 65

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

Answer 66

omg im blind

Answer 67

:3

Answer 68

(-10h-h^2)/h

Answer 69

can i simplify more than that?

Answer 70

yah you can divide an h out of every term.

Answer 71

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

Answer 72

-10-h is the final answer?

Answer 73

yay good job \c:/

Answer 74

yay thank you!

Answer 75

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

Answer 76

sure

Answer 77

working on b?

Answer 78

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

Answer 79

shouldnt it be -3?

Answer 80

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

Answer 81

1?

Answer 82

yup!

Answer 83

omg!!! ty

Answer 84

u make it look so easy

Answer 85

could you help me with part e) that looks complicated

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

5?

Answer 90

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

Answer 91

f(5)=?

:o

Answer 92

f(5)=4

Answer 93

sqrt(4-4)

Answer 94

undefined?

Answer 95

Mmm sqrt0 = 0

cuz 0*0=0

Answer 96

Or better said: 0^2=0

Answer 97

it says 0 is not the correct answer

Answer 98

Hmmm :\