All Answers 358524 Answers

Alternatively, avoiding the x squared term,

log(x/10) = log(10/x) + 2,

log(x) - log(10) = log(10) - log(x) + 2,

2log(x) = 2log(10) + 2

= 2 + 2

= 4,

log(x) = 2,

x = 100.

I understand what you are saying Yura but I would love to be teaching you face to face right now.

I know you are expected to memorize the formula and I understand that formula are nice and comfortable and if you memorize them they work really well.

I too learned formula for this before I really understood what was happening.

BUT if you understood what was happening you would not need the formula - the method would just be obvious.

I know there is very little time for anything when you studying conscienciously full time but try to make time to understand graphs.  If you understand them then many other things become so much easier.  Many things just become obvious!  

Here is an introduction to derivatives that I'd like you to take a look at.

It is not exactly what I was looking for but it may serve you well anyway.  :/

Sun  7/2/16

1)  How many tickets??....good reasoning problem answered by Dragonslayer and a guest  :

https://web2.0calc.com/questions/ariana-akizian-has-a-question_1

2)  Arrows on the target  .....asked by DS......answered by me and Solveit....

https://web2.0calc.com/questions/probability-math-question

3)  A good number problem posted by Shades, answered by a guest : https://web2.0calc.com/questions/number-problem_1

4)  Finding the diagonal of a sqaure.....I really liked this one that rarinstraw submitted!!!.....I did it one way......heureka did it entirely differently [ I don't understand what he did, but I'm hoping he'll give me some guidance]

http://web2.0calc.com/questions/another-geometry-question_1

5)  Finding an equation of a circle.....geno did it one way....I did it another...

http://web2.0calc.com/questions/writing-an-equation

6)  An interesting geometry question from chilledz3non:

http://web2.0calc.com/questions/help-with-algebra-2-problem

This one requires some knowledge of Euclid's Elements.....I believe I got the correct answer....

7)  Some head-scratchers asked by EinsteinJr  and answered by Haley, guest and me : https://web2.0calc.com/questions/a-few-problems          [one is still unanswered ]

@@ What is Happening?  [Wrap4]   Sat 6/2/16   Sydney,  Australia Time 5:40pm   ♪ ♫

Hi everyone,

As most of you would be aware, I have been away and I have not been very active on the forum for the last 2 or 3 weeks.  I have not been collecting names of answerers but I greatly appreciate all the help that every answerer gives

What is happening:

1)  26th January was not just Australia Day AND not just Republic Day in India.

      26th January is also

                 Birthday of Camelot's Valiant Protector,  TitaniumRome!!  

                    Happy belated birthday TR.     I hope it was a great day!  

2)   What else?   How should I know, I've not been here   LOL    

I did notice something this morning

There used to be 5 forums here: English, German, Spanish, French and Russian.

I knew that the Russian forum, which was not being used, was removed a while back but I noticed this morning that it has been replaced by HINDI     

Hindi is one of the official languages of the Union of India and it is the first language for 180 million people.

My goodness that is a lot of people !!   Australia only has 24 million people!   I digress........  

We have many Indian members and guests on the forum.  Perhaps some of those can get some good use from the new Hindi Maths Forum.  Thanks Mr. Massow for organising this forum for the 180 million potential users. It will be great to see it come to life.   

You can click into any of the forums from the bottom of most screens.  

Just click into the one that you want  (This is what it looks like)    

Interest Posts: 

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.

I have not been around and I have not sure I have not seen a great many fabulous questions and answers, these are just a few of the good ones that I have, by chance, seen.  

Communications and Fun:

1)   Some wonderful recognition for the forums hard working answerers.   Thanks Guest :)

Thanks Coldplay   

Yes, you made it in the BIG TIME rarinstraw   CONGRATULATIONS           

AND Most of all - Happy belated birthday TitaniumRome.   

Divide both sides by 30

your question does not make sence :(

It was on my watch list for a very lonnnnng time

after Mosscow built the new Camelot "stadium" and replaced the old one, i went through my entire watchlist and decided out of all the old ones, this was one great memory I would keep, because it best defined how great everything was back then and how everyone had fun (and that was also when I didn't or anyone else made others mad), it was a memory overall in my opinion to save for all time

CPhill: I agree with you. However, the software spat that out  qualifying it thus: "x=-100 [assuming a complex-valued logarithm]", which I ignored!.

Gee Titanium,

You found a great old thread there  :))

Sorry, guest......

x cannot be -100  because it would mean that we would be taking the log of a negative number on both sides in the original equation....!!!

log(x/10) = log(10/x)+2           [   2 = log (100) ]

log(x/10)  = log (10/x) + log (100)

log(x / 10)  = log ( 10/x * 100)

log(x/10)  = log ( 1000/x )     and......equating logs, we have

x/10  = 1000/x       cross-multiply

x^2  = 10000

x = 100

Solve for x: (log(x/10))/(log(10)) = 2+(log(10/x))/(log(10)) Subtract 2+(log(10/x))/(log(10)) from both sides: -2-(log(10/x))/(log(10))+(log(x/10))/(log(10)) = 0 Bring -2-(log(10/x))/(log(10))+(log(x/10))/(log(10)) together using the common denominator log(10): -(2 log(10)+log(10/x)-log(x/10))/(log(10)) = 0 Multiply both sides by -log(10): 2 log(10)+log(10/x)-log(x/10) = 0 2 log(10)+log(10/x)-log(x/10) = log(100)+2 log(10/x) = log((100 10 10)/(x x)) = log(10000/x^2): log(10000/x^2) = 0 Cancel logarithms by taking exp of both sides: 10000/x^2 = 1 Take the reciprocal of both sides: x^2/10000 = 1 Multiply both sides by 10000: x^2 = 10000 Take the square root of both sides: Answer: | | x = 100   or   x = -100

Thanks, Melody.......!!!  ......here's another way to see this...

This must be a linear equation if the derivative is just some integer at all x values...so....

At  0, f(x)  = -5

So....at f(2).......x  has changed by 2  units

Thus, using the formula for slope :

 [ f(2) - (-5) ] / [ 2  - 0]  <=  3     simplify

[ f(2)  + 5]  / 2  <= 3     multiply both sides by 2

f(2) + 5 <= 6      subtract 5 from both sides

f(2)  <=  1

Thus f(2) cannot  = 2   because this is greater than 1

Let  one number  = -15    and the other - 2

So

  -15 *  - 2   = 30

And

-15 +  -2   = -17

ok then, if the gradient was equal to 3 then f(2)=1

if the gradient is less than 3 then f(2) must be less than 1 (because it is not so steep anywhere)

so f(2) cannot be more than 1.    Only one of those points is more than 1.   :)

im sorry its suppose to be f'(x) lesser or equal to 3

What is the equation?

Problem 2

Put 3 jewels on one side and 3 on the other :

1st scenario - they balance

Take them all off

Pick any two of the remaining three jewels

Put them on the scale.....if they balance....the remaining one is the real one

If they don't balance......the real one has to be the heavier

Result......real one is found after just two weighings

2nd scenario - they don't balance and one side is heavier

Take the three jewels from the heavier side

Pick any two of these

Put them on the scale.....if they balance....the remaining one is the real one

If they don't balance......the real one has to be the heavier

Result......real one is found after just two weighings

let f be a differentiable function such that f(0)=-5 and f'(x)= 3 for all x, but which value is not possible for f(2)?

what you must remember is that f'(x) is the gradient of the tangent to the curve.  If this gradient is any constant then the curve MUST be a line.

SO

This is a line with a gradient ot 3 and a y intercept of -5

the equation has to be

f(x)=3x-5

f(2)=2*3-5 = 1

No other value is possible.

im guessing its gonna be 2 b/c maxmium increase is 6, and 2 is not in its range, is this right 

Trapezoid: b:x B:x+2 h:3

Area of the Trapezoid =   (3/2)[x + x + 2]  = (3/2)[2x + 2]   = (3/2)*2[x + 1]  = 3[x + 1]

Parallelogram: b:x+4 h:2

Area of the Parallelogram  =  [x + 4]*2

So....setting these equal, we have......

3*[x + 1]  =  [x + 4] *2

3x + 3  = 2x + 8       subtract 2x, 3 from both sides

x  =  5

The area is 18 sq units for each figure.....