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By a log property we can write

Log 4x + Log 2x = Log (8x^2)       So this gives us

Log(8x^2) - 4 = 0      Subtract 4 from both sides

Log (8x^2) = 4          Which says that, exponentially

10^4 =  8x^2       

10,000 = 8x^2           Divide by 8 on both sides

1250 = x^2                Take the square root of both sides

±√(1250) = x =       ± 25√(2).....we must reject the negative root because it would mean that we would have the log of a negative number in the original equation, and that's undefined.

Let h be the length of the hypoteneuse. Then by the Pythagorean Theorem, we have

(h-4)^2 +  (h-8)^2 = h^2       Expanding the terms, we have

h^2 - 8h + 16 + h^2 - 16h + 64 = h^2        And simplifying

2h^2  - 24h + 80 = h^2         Subtract h^2 from both sides

h^2 - 24h + 80 = 0               And factoring, we have

(h -20) (h-4) = 0             Reject h =4, since it would make one side 0 and the other -4

So the hypoteneuse is 20 and, one side = (20-4) = 16, and the other side = (20 - 8) = 12

pythagorean theory. not an 100% sure it will work but try it out. if not best way= regoogle it.

z= 1

A= -2

B= -1

That would be a horizontal line melody 

I'd say $$[(x,y)\in\mathbb{R^2}:x=c_1,c_2 \leq y \leq c_3]$$

Which stands for, x and y can both be any real number for which x is equal to c1 and y varies between c2 and c3

 It may seem a bit weird to say x is any real number for which it is equal to c1, but it's only an indicator for the type of values we are looking at (integers, complex numbers, postive values, etc.)

Reinout 

download what?

Allright so I found some answers online and I'm going to combine them to give you a proper answer;

'How many solutions total are there to x1 + x2 + x3 + x4 + x5 = 21 ? 

Imagine you have 21 "objects" which we can represent with the letter o: 

      o o o o o o o o o o o o o o o o o o o o o 

I think that's 21 of them. Now, how do we divide them up into 5 non-negative values? Just split them up into five sections. The way to do that is just like this: 

      o o o o | o o o o | o o o o o o |o o o o o | o o 

That corresponds to the solution x1=4, x2=4, x3=6, x4=5, x5=2 

Now how many ways can you do that? There are 20 viable places for the separator marks -- anywhere between the 21 o's. So you pick a place for four of them -- 20C4. 

But those are the solutions with positive values -- you want non-negative values. 

How can we compensate for that? Well, we can reduce this to a different problem. 

If I want to solve things with non-negative values x1,x2,..,x5, I can simply consider some positive number y1 and then let x1=y1-1. That will produce all non-negative possibilities. 

So you just solve: 

      x1 + x2 + x3 + x4 + x5 = 21 
      (y1 - 1) + (y2 - 1) + (y3 - 1) + (y4 - 1) + (y5 - 1) = 21 
      y1 + y2 + y3 + y4 + y5 = 26 

So now y1,y2,..,y5 are positive solutions -- we've just "transformed" the problem into a problem we know the answer to, just with different parameters. The answer is 25C4 instead of 20C4. '

could u just tell me the reason for ur steps that u are doing ! im sure after this ill understand this ques. very nicely 

We are solving the equation f(x)=0. Graphically the roots (solutions) of the equation are the x co-ordinates of the points where the y=f(x) crosses the x axis, i.e. the points where y = 0.

The Newton-Raphson method starts with a point x0 close to a root, takes the tangent to the curve for that value of x, (the tangent at A in the sketch below) and then calculates the x co-ordinate of its intersection with the x axis, (x1 in the sketch below).  (All done algebraically of course). Hopefully x1 will be a better approximation to the root  than was x0. The process is then repeated using x1 as the new starting point to find an x2 then again to find an x3 and so on until converged to whatever degree of accuracy is required.

The method can go awry if the gradient of the tangent is small, it's possible to be kicked a long way (possibly irretrievably) from the original point. In the sketch below, for example, suppose x0 were further to the right so that the tangent to the curve was constructed at B. The next approximation could be way off to the left or the right depending on whether B is to the left or right of the maximima. Either way the next 'approximation' to the root will be useless. On paper, the problem manifests itself with a small value for f '(x) and a correspondingly large correction term f(x)/f '(x).

I'm still thinking about the simplest way to solve this, but I think these are seperate questions Alan.Since the values need to be nonnegative integers there's a limited amount of solutions for each case...

If your not yet studying logarithms, you can off course also make a table for some values of x.

You'd find that the 3000 is somewhere between x=18 and x=19. 

All right, so your question is;

$$1200 \times (1.05)^x = 3000$$

Do you know what a logarithm is?

Basically if we have

$$a^b = c$$

then

$$log_ac = b$$

What a wonderful explanation CPhill 

I have nothing to add

Do you mean this bit?

$$\begin{array}{rll}
\frac{3500}{T}&=&\left(\frac{3000}{T}+50\right)\\\\
T\times\frac{3500}{T}&=&T\times\left(\frac{3000}{T}+50\right)\\\\

\end{array}$$

I multipled both sides by T so that I could get the T off the bottom.

See, the next line just becomes this;

$$\begin{array}{rll}

3500&=&3000+50T\\\\

\end{array}$$

Because the Ts cancelled out

I'll put in a bit more working to show you.

$$\begin{array}{rll}
\frac{3500}{T}&=&\left(\frac{3000}{T}+50\right)\\\\
T\times\frac{3500}{T}&=&T\times\left(\frac{3000}{T}+50\right)\\\\
\frac{3500\times T}{T}&=&\frac{3000\times T}{T}+50\times T\\\\

3500&=&3000+50T\\\\
500&=&50T\\\\
T&=&10
\end{array}$$

If you still do not understand let me know.   Also tell me if it is still the same spot that is causing you problems because I already have ideas on how I can help you understand.  Just think about it for a little while first though.   

That is great Rosala,

I'd like to chat with you. I'll send you a message.  

thats nice melody u could understand his question . thumbs up from me 

ive read it 

melody why did u do time into 3500 upon time ?

Your constraints are inconsistent.

a) constrains x1 to be greater than zero; c) allows it to be zero.

b) constrains x2 to be greater than 1; d) allows it to be 1.

Which constraints apply?

@@ End of Day Wrap - Saturday 24/5/14   Sydney, Australia   Time 21:30

Hi everyone,  

I trust that you are all enjoying your weekend!  It has been a fabulously sunny day in Sydney!

Great answers were given today by Takahiro Maeda, CPhill, Rosala, Alan, Rom, reinout-g, rag3phone, zegroes, nadiyah and Bertie.  Thank you, you are all wonderful.

I haven't heard from Andre Massow but I saw no more evidence of the hacker today so presumably Andre has taken care of the problem.  Only Chris and I were complaining yesterday.  It was driving us nuts.  I am wondering if we were the only ones affected. Doesn't matter, hopefully the problem is finished with.

Here are some posts that may be of interest.

Speed questions.  Good (High level) junior high problems.  Students often find questions relating to speed very difficult.  I think that they are often quite interesting.  Rosala presented a few today.

Can anyone answer anonymous's question?

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"But I see that it can also give wrong estimates sometimes

Can someone inform me about these cases? please :D"

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Thanks you Bertie.    

Hi Stacy

You need to use the quadratic formula.