imaginator
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Staring at the window and experiencing the nature is better than studying Human physiology for me...
Posts: 62
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Post by imaginator on Mar 26, 2015 6:25:19 GMT
One fourth of one third of two fifth of a number is 15. What will be 60% of that number? It's getting trickier as much as I'm trying...
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Post by airbournebearcarcass on Mar 26, 2015 9:55:52 GMT
I am now acutely aware of the tedious and terribly confusing formatting of my Logarithmic equations, for I am having a hard time reading people’s workings I believe it is very likely that people have misread my equations into something unsolvable. From what I can comprehend so far, only dogfish 44 correctly simplified the second equation. 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] 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) 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] There two more important change of base rules to consider log (a^r)X=(1/r)(log aX) note: '^' means 'to the power of' (log ca)(log bc)=log ba To avoid frustrating people any further here is the full solution, it is handwritten to make no room for confusion, bear with my handwriting. If your workings are actually correct but I failed to acknowledge then you have my apology. The key to solving this question is to simplify the Logarithms into constants before converting them to index form, X and Y will be in the indexes and once you reached there it should not be impossible to solve even without calculator. The equations are formatted a bit differently when handwritten, due to the computer's inability to type out fractions and surds. The fractions and surds are expressed by the '÷' symbol and as decimal indexes respectively when I typed my equations on computer. In my handwritten solution I also carelessly used 'x' as the multiplication symbol, please do not confuse them with X the unknown. If my equations typed on the computer are actually different from what I wrote in my solution then you have my deepest apology for wasting your time and frustrating you. They are split into four parts I know it is a really lame answer. Ichthyophobia you win for getting the correct answer by guessing.
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dogfish44
Junior Member
You are now aware that you are breathing manually!
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Post by dogfish44 on Mar 26, 2015 12:42:02 GMT
... I already mentioned these, just by a different form As for the final simplification, TT got to the end of the Y eq. as well. I just mindblanked on switching to another power of Y (I blame 4 AM).
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Post by ichthyophobia on Mar 26, 2015 15:28:31 GMT
Imaginator, the answer to yours is 270.
14*4 = 60 60*3 = 180 180/2*5 = 450 0.6 * 450 = 270
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imaginator
Junior Member
Staring at the window and experiencing the nature is better than studying Human physiology for me...
Posts: 62
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Post by imaginator on Mar 27, 2015 6:52:28 GMT
oooh..thank you Ichthy..
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Post by airbournebearcarcass on Mar 30, 2015 9:47:44 GMT
... I already mentioned these, just by a different form As for the final simplification, TT got to the end of the Y eq. as well. I just mindblanked on switching to another power of Y (I blame 4 AM). Sorry my bad A more sensible question than my last The relationship between the number of times Aoko chased Kaito with a mob in a week and the number of times Kaito flipped up Aoko's skirt in a week is represented by the equation y=bx c. When the equation is plotted on a graph, it is a curve which passes though two points whose coordinates are (16,6) and (81,9). b and c are constants, find the value of b and c. Note: In a coordinate, the value of the x-axis is represented first
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yesterday
New Member
"Don't shoot! I'm Canadian!" - X Company S2E9
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Post by yesterday on Feb 18, 2016 0:27:58 GMT
Okay, so one of the guys in my woodworking class was studying for his math midterm and showed me this question. Can anyone figure out the length of AB? 'Cause I can't.
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Post by midnight on Feb 18, 2016 21:41:37 GMT
yesterday This is frustrating me in more ways than it should, more than anything because I remember having solved it before and can't remember how. The only thing I remember is the name of the technical's drawing version of this exercise, called 'trazar dos tangentes internas comunes a dos circunferencias de radios diferentes' which would translate to something like 'two common internal tangents to two circles of different radius'. I know it's not much but this problem's answer is definitely online, it should have a similar title if you want to find out the answer fast, that is. The only other thing I can add, not sure if it's much help, is that segment AB makes 90º with the radius of 24. The same with the other circle
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dogfish44
Junior Member
You are now aware that you are breathing manually!
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Post by dogfish44 on Feb 18, 2016 22:06:54 GMT
Ooh, that looks fun. Stream of consciounce, hiyoo! {Spoiler}Let's describe the centre of the left circle as point C, and the centre of the right as D. Let's draw line CD, which we know to be length 70 cm. Let's mark where the line going down from D and the circle meet as E.
Triangle ABC and triangle BDE are basically the same, but scaled differently and rotated around B. This means that the ratio between lengths is the same. The ratio can easily be found using lines AC and DE, which gives a ratio of 4:3.
Now we know that, we can figure out BC and BD easily. BC and BD total 70, and are split 4:3. BC is 40cm, and BD is 30cm.
Now this is the bit where I assume that AB runs at a tangent to that circle, since otherwise I can't calc this out. If it is, then ABC is a right angled triangle, and Pythagoras gives AB as √(1600 - 576), which comes to a rather neat 32cm.
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yesterday
New Member
"Don't shoot! I'm Canadian!" - X Company S2E9
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Post by yesterday on Feb 19, 2016 1:02:35 GMT
*gasps* I knew you guys could do it! Poirot Cafe for the win!
Thanks so much midnight and dogfish! I think my teacher skipped circle geometry last year 'cause we had a teacher strike, so thanks a bunch for solving it. (And with your explanation too!) I feel slightly smarter now.
And now I gotta study for my physics midterm.
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Post by doctorpeggy on Feb 12, 2018 13:20:39 GMT
I was practicing for exams, and found this question that I couldn't understand for the life of me. I can't seem to put in the image, so here's a link to it: drive.google.com/file/d/19gsMiB5oqLLTWd0NUHhPOm3M4UEZwASu/viewI'm stuck on question 2, and every time I think I've found something for part a, part b throws me off. Can anyone help?
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