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Post by Mikauzoran on Dec 26, 2014 14:21:57 GMT
I got the same answer as Natsumi. {Answer and My Work} I got that the kid is 40 and the father is 60.
Work:
x=kid's age when kid was younger. y= father's age when kid was younger and kid's age now z= father's age now
y+z=100 y=2x z=x+y
Now if you plug those values back into the original y+z=100, you get: 2x+x+2x=100 divide each side by 5, and x=20. You can plug 20 into the equations above to get their respective ages now: 40 and 60.
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Kaggami
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I´m freezing in my own house... winter, no one invited you yet!
Posts: 527
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Post by Kaggami on Dec 26, 2014 15:34:02 GMT
It´s right! XD
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Post by StarlightDragons on Dec 26, 2014 16:08:30 GMT
yay!
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Post by Nikudou Natsumi on Dec 26, 2014 16:21:38 GMT
Yes!
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Post by Mikauzoran on Dec 26, 2014 20:27:45 GMT
Huzzah!!! ^.^ These are so fun now that I don't have to actually do them for a grade. It's good to stretch your mental muscles (or shall I say little grey cells? We are Poirot Cafe, after all).
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Post by airbournebearcarcass on Mar 25, 2015 13:13:24 GMT
Kaito has (XY)! number of pink dove(s)
log16[(log54)(log210)÷(log25100.5)](2 log26 - log29)Y=X
log{[(log54)(log210)]÷log25100.5-(log35)(log527)}[(log464)(2 log26 - log29)(25X-1)-ln e]=Y
In case there is any confusion regarding the second equation, {[(log54)(log210)]÷log25100.5-(log35)(log527)} is the base of the logarithm.
Solve the equations to find X and Y, and hence find the number of pink dove(s) Kaito keep.
What have I done...?
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Post by Nikudou Natsumi on Mar 25, 2015 13:57:53 GMT
What's with all these logs?! I only just learned logs . . . And I've yet to learn ln things because our lesson got cut short . . . But I'll learn more about them today, so I'll try this later!
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Post by StarlightDragons on Mar 25, 2015 16:24:33 GMT
....why nobody tell anyone, I'm going to try and piece this out. >.>
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dogfish44
Junior Member
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Post by dogfish44 on Mar 25, 2015 16:34:11 GMT
Logs. I hate logs.
For the proverbial record, ln = loge. It comes up an awful lot, so we give it it's own shortenint.
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Post by StarlightDragons on Mar 25, 2015 17:13:30 GMT
Wait wait, for the first equation ( for x ) is that last part multiplied at the log or is it part of the log's 'insides', since it's not part of the parenthesis?
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Post by Nikudou Natsumi on Mar 25, 2015 18:24:13 GMT
I was doing fine . . . but then y was in the x equation and x was in the y equation -.- And now I don't have time to solve from there.
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TinyTantei
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Post by TinyTantei on Mar 25, 2015 20:07:01 GMT
Man, I have to hand it to you, this one's a toughie...I've got it simplified quite a ways, but nothing too concrete yet. Still chugging away at it, though!
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Post by Nikudou Natsumi on Mar 25, 2015 23:09:24 GMT
Okay, this is what I have so far, but I'm kinda stuck right now. {Spoiler: Click to view/hide} Y = log5(6(25^(log168*2^Y)-1)-1) X = (log168*2^(6(25^X-1)-1) (in the Y equation, the first -1 is still part of 25's exponent) But I don't know how to get the variable exponents down. I know you can do it with logarithms, but I'd have to do it to the other side, and I don't know how to do that when the exponent is layered in a bunch of other processes.
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TinyTantei
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Post by TinyTantei on Mar 26, 2015 1:16:35 GMT
Mmm, I'm on the last stretch of the problem myself, the process I took was really long so I won't put it here unless somebody really wants to see it all written out, but: {Constants}Okay, so there are two really really big numbers that are actually constants so they're not so scary once you break them down? Let's see, which ones did I use variables for...
n = log(5)4*log(2)10/log(25)10^0.5 = 8
log(16)n = 3/4
(There was a lot of change of base involved with finding that answer), and
m = log(3)5*log(5)27 = 3
There are a few others, but those are the main two I saved til the end. And then I've basically just boiled those two equations down to: {Work}X = (3/4)2^Y
which can then be changed to
Y = log(2)4x/3
And then for the Y equation, I have
Y = log(5)[(6*25^(X-1))-1]
which you can equate to each other to find out that
log(2)4x/3 = log(5)[(6*25^(X-1))-1] ... And now that's where I'm stuck, because I don't think change of base works in this scenario (unless it does and I'm just being a doof).
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dogfish44
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Post by dogfish44 on Mar 26, 2015 2:48:01 GMT
Like Tiny I'm stuck at the same equality. My full workings and notes. log 16[(log 54)(log 210)÷(log 2510 0.5)](2 log 26 - log 29) Y=X log {[(log54)(log210)]÷log25100.5-(log35)(log527)}[(log 464)(2 log 26 - log 29)(25 X-1)-ln e]=Y A few base rules I'll probably use as I go, for reference: log ab = log xb / log xa log aXY = log aX + log aY log aX/Y = log aX - log aY log aX n = nlog aX log aa = 1 log(X) = log 10(X) [Clarity] {Equation 1}X = log16[(log54)(log210)÷(log25100.5)](2 log26 - log29)Y
Two sections to this. First of all, let's consider the second section, (2 log26 - log29)Y
2 log26 becomes log262, or log236
Both answers are in base 2. Ergo, we can subtract them by dividing their arguments. This gives log24
From here, we see that the argument 4 - or 22. From this, we can get 2log22
Since the base and the argument are the same, the log is equal to 1. So we have 2(1), or just 2.
Ergo, (2 log26 - log29)Y = (2)Y
X = log16[(log54)(log210)÷(log25100.5)](2)Y
Next, we look at the part log16[(log54)(log210)÷(log25100.5)]
The first thing to notice is that the denominator log has a power in it. We can simply multiply this out. log16[(log54)(log210)÷(0.5 log2510)]
Let's try to simplify the final expression. log54 can be written as 2log52, and we can divide by 0.5 on both parts. log16[4(log52)(log210)÷(log2510)]
From this, we can attempt to calculate. Multiplication will probably be easier. Let's consider: (log52)(log210)
One way to make sure multiplications go easily is to attempt to get similar numbers. For instance, (log210) can be written as ((log25) + (log22)), or more elegantly as (log25 + 1).
So now to multiply (log25 + 1) and (log52)
We can use our first rule to make the powers equal. (log52) = (log2/log5) and (log25 + 1) = (log5/log2 + 1)
(log2/log5)(log5/log2 + 1) = (log2)(log5)/(log5)(log2) + log2/log5. (log2)(log5)/(log5)(log2) simplifies down to 1.
In short, (log52)(log210)= 1 + log2/log5
Subbing that in, log16[4(1 + log2/log5)÷(log2510)]
Now to consider (1 + log2/log5)÷(log2510)
First, we will need to convert log2510 to base 10. This is equal to log10/log25 (1 + log2/log5)÷(log10/log25)
We can change that to a multiplication by inverting one side. (1 + log2/log5)*(log25/log10)
The "1 +" is annoying. Temporarily, let's change back log2/log5 to log52 If we convert 1 to log55, we can add it onto log52 for a single log510. We can now convert back to base 10, which will be log10/log5, and sub it back in.
(log10/log5)*(log25/log10)
We can simplify log25 (log10/log5)*2(log5/log10)
log10s cancel, log5s cancel, we ultimately end up with 2.
(1 + log2/log5)÷(log2510) = 1
Subbing that in, log16[4(2)]
And subbing that into X outright, X = log16[4(2)](2)Y X = log16[8](2)Y
The last thing we'd ideally do is get rid of the log. Since 8 = 23 and 16 = 24, we might be able to simplify this down quickly.
We can say that log16[8] is equal to log8/log16. From this, we can say that log8n/log16 = 1, and that log8n = log16
Or, we can say that log23n = log24
n must then 4/3. We can multiply out this simple power to get that 4/3log23 = log24
Or that 4/3 * log8 = log16. Or that log8 = 3/4 * log16. OR, more notably, that log8/log16 = 3/4.
We can sub that in now, so, X = 3/4 * 2Y
This also gives us a chance to create a value for Y 4X/3 = 2Y --> Y = log2(4X/3) {Equation 2}Y = log{[(log54)(log210)]÷log25100.5-(log35)(log527)}[(log464)(2 log26 - log29)(25X-1)-ln e]
Let's deal with simple stuff. ln e = 1, log464 = 3, (2 log26 - log29) = 2 (See above). Let's shrink the equation by subbing these in.
Y = log{[(log54)(log210)]÷log25100.5-(log35)(log527)}[3*2*(25X-1)-1]
We also solved (log54)(log210)÷(log25100.5). That bit equals 4(2). Or 8. We can also simplify 3*2. I hope.
Y = log{8-(log35)(log527)}[6(25X-1)-1]
We can now look at (log35)(log527). Let's start by acknowledging 27 is 33, so we can translate this to (log35)(3log53).
Transferring to base 10 this is log5/log3 * 3(log3/log5). log3s and log5s cancel, leaving just 3. Or, (log35)(log527) = 3. Subbing that in,
Y = log{8-3}[6(25X-1)-1] Y = log5[6(25X-1)-1]
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