anonymous
  • anonymous
Write the expression below so that only a single logarithm or exponential function appears. (1/2)ln(z) − ln(5 + x) − 4 ln(y)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zepdrix
  • 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
  • anonymous
ln(z)^(1/2)
zepdrix
  • 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
  • zepdrix
We can do the same with the 4, yes?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ln(z^(1/2))-ln(5-x)-ln(y^4)
zepdrix
  • zepdrix
\[\Large \ln \left(z^{1/2}\right) − \ln(5 + x) − \ln\left(y^4\right)\]Ok cool.
zepdrix
  • 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
  • anonymous
I get this ln((z^(1/2))/((5+x)(y^4)))
zepdrix
  • 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
  • anonymous
yay thank you:)
anonymous
  • anonymous
hey zep could u help me with another problem.. i dont know what im doing wrong
zepdrix
  • zepdrix
sure
anonymous
  • 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
  • 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
  • anonymous
for part a) I'm getting (1.7)^tM and it says is wrong
zepdrix
  • zepdrix
Is the M in the exponent? I can't tell with the way you wrote it :3
anonymous
  • anonymous
no multiplying
anonymous
  • anonymous
(1.7^t)*M
zepdrix
  • 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
  • zepdrix
\[\Large 1.7=b^{10}\]
zepdrix
  • zepdrix
\[\Large b\approx 1.054\]
zepdrix
  • zepdrix
Do you have a way to check this? To make sure I'm not doing something really stupid :) lol
anonymous
  • anonymous
well i need a formula for part a
anonymous
  • anonymous
and i get (1.7^t)M and no i dont have a way to check it :(
zepdrix
  • 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
  • zepdrix
^ Does this way make sense? :o How did you come up with b=1.7?
anonymous
  • anonymous
yup that worked!
anonymous
  • anonymous
you are amazing!
zepdrix
  • zepdrix
yay team \c:/
anonymous
  • anonymous
hey for part b do i just plug 10 for t?
anonymous
  • anonymous
and solve for M?
zepdrix
  • 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
  • zepdrix
\[\Large 10M
zepdrix
  • 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
  • anonymous
ok let me try
anonymous
  • anonymous
t is about 43.3936
anonymous
  • anonymous
so 4 days?
anonymous
  • anonymous
44
zepdrix
  • zepdrix
ya that sounds right! :)
anonymous
  • anonymous
says its wrong :(
zepdrix
  • zepdrix
Hmm thinkinggg
zepdrix
  • 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
  • anonymous
that worked!!!!
anonymous
  • anonymous
thank you sooo much!
zepdrix
  • zepdrix
Oh cool! :) Strange wording on the problem. heh
zepdrix
  • 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
  • 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
  • anonymous
yea im having trouble with this part
zepdrix
  • 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
  • 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
  • anonymous
ok so now i solve for h
zepdrix
  • 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
  • zepdrix
Not solve for h, we're just trying to simplify the expression.
zepdrix
  • zepdrix
Do a bunch of multiplication, expand out the binomial with the square on it, and simplify! :D
zepdrix
  • zepdrix
Remember how to expand this out? \(\Large (3+h)^2\)
anonymous
  • anonymous
omg this is going to be a pain... let me try this lol
anonymous
  • anonymous
i get (-36-3h+h^2)/h
anonymous
  • anonymous
not sure on that
zepdrix
  • 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
  • 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
  • zepdrix
See any mistakes you may have made? :o
anonymous
  • anonymous
-12-4h-6h-h^2?
anonymous
  • anonymous
-12-10h-h^2?
zepdrix
  • 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
  • anonymous
omg im blind
zepdrix
  • zepdrix
:3
anonymous
  • anonymous
(-10h-h^2)/h
anonymous
  • anonymous
can i simplify more than that?
zepdrix
  • zepdrix
yah you can divide an h out of every term.
zepdrix
  • 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
  • anonymous
-10-h is the final answer?
zepdrix
  • zepdrix
yay good job \c:/
anonymous
  • anonymous
yay thank you!
anonymous
  • anonymous
I have a similar question.. could u set it up for me and let me try to do it?
zepdrix
  • zepdrix
sure
zepdrix
  • zepdrix
working on b?
zepdrix
  • 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
  • anonymous
shouldnt it be -3?
zepdrix
  • 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
  • anonymous
1?
zepdrix
  • zepdrix
yup!
anonymous
  • anonymous
omg!!! ty
anonymous
  • anonymous
u make it look so easy
anonymous
  • anonymous
could you help me with part e) that looks complicated
zepdrix
  • zepdrix
\[\Large h(\color{#3399AA}{x})=\sqrt{\color{#3399AA}{x}^2-f(g(\color{#3399AA}{x}))}\]
zepdrix
  • zepdrix
\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(g(\color{#3399AA}{2}))}\]
zepdrix
  • zepdrix
Again, work from the inside out,\[\Large g(2)=?\]
anonymous
  • anonymous
5?
zepdrix
  • zepdrix
\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(5)}\]
zepdrix
  • zepdrix
f(5)=? :o
anonymous
  • anonymous
f(5)=4
anonymous
  • anonymous
sqrt(4-4)
anonymous
  • anonymous
undefined?
zepdrix
  • zepdrix
Mmm sqrt0 = 0 cuz 0*0=0
zepdrix
  • zepdrix
Or better said: 0^2=0
anonymous
  • anonymous
it says 0 is not the correct answer
zepdrix
  • zepdrix
Hmmm :\
anonymous
  • anonymous
what did we do wrong :(
zepdrix
  • zepdrix
Did you try undefined yet? :O Seems like that shouldn't be the answer, but whatev, it's worth trying.
anonymous
  • anonymous
no undefined is not an answer either :(
zepdrix
  • zepdrix
Oh I see, that is not the composition operator in part e, my bad. That's multiplication I think.
zepdrix
  • 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
  • zepdrix
Umm let's try it and see what happens.
zepdrix
  • zepdrix
\[\Large h(\color{#3399AA}{x})=\sqrt{\color{#3399AA}{x}^2-f(\color{#3399AA}{x})\cdot g(\color{#3399AA}{x})}\]
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
\[\Large h(\color{#3399AA}{2})=\sqrt{\color{#3399AA}{2}^2-f(\color{#3399AA}{2})\cdot g(\color{#3399AA}{2})}\]
zepdrix
  • zepdrix
We get this, yes?\[\Large h(2)=\sqrt{2^2-(-1)\cdot(5)}\]
anonymous
  • anonymous
sqrt(-1)?
anonymous
  • anonymous
or sqrt(9)
zepdrix
  • zepdrix
ya sqrt9 i think :L
anonymous
  • anonymous
yay that worked!!!!
zepdrix
  • zepdrix
yay team!
anonymous
  • anonymous
:)
anonymous
  • anonymous
hey zep do u have time for one more question

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