All Answers 358087 Answers

$${{\mathtt{199}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{34}}}}\right)} = {\mathtt{1.168\: \!458\: \!590\: \!290\: \!93}}$$

Hi Bane of Dracula and welcome to the forum :)

Do you know how many friends John is sharing with or is that meant to be a part of the confusion?

Here's a more general approach (there are an infinite number of solutions, but only two real ones):

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Use common multiples although it is the most retarded way,P5 pupils use this method

Multiples of 8(-5): 3 11 19 27 35 43

Multiples of 6(+19): 25 31 37 43

Ans:43

Thanks Alan,

I tried to do it this way with the help of Wolfram|Alpha but I didn't get an answer.

I might have been doing it wrong.

Also,

When you did it with Mathcad, it doesn't seem to have thrown up my answer.  Am I wrong?  Why was that?

This is how I found it (with the help of Mathcad):

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@@ End of Day Wrap  Mon 10/8/15     Sydney, Australia Time   8:40pm    ♪ ♫

Good evening all,

Our iillustrious answerers are Dbg4thebest, Geno3141, Rubenhh, Alan, Heureka, CPhill, jdh3010 and ShimmyCocopang.  Thanks all.  

30% of 120 is 36, so Tom originally had 36 red marbles.  If his father gave him n red marbles then

36+n = (120 + n)*40/100   Multiply through by 10 and simplify:

360 + 10n = 480 + 4n   Subtract 4n from both sides

360 + 6n = 480   Subtract 360 from both sides

6n = 120  Divide both sides by 6

n = 20

Tom's father gave him 20 red marbles.

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Thanks Alan, now it works.

I have edited my check of your answer.

How did you get your answer and are you sure that there are no others?

Your fourth line should have (5x² + 49x - 10)i Melody, not (x² + 49x - 10)i

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Why have you asked this exact same question twice ?

Hi Dbg,  

Jdh3010's graph is NOT a function because it does not pass the vertical line test.

When you draw any vertical line through a function it can cross the fuction in AT MOST one place.  

This graph would often have two crossing points :)

This means that some x values have more than one y value.      

If a graph or an equation is a function it means that no x value can have more than one y value.  

2*5-1-0=9  that is one answer.

150-50=100

100/2=50

he had 50 left

Taking logs, the equation can be written as ,

and this requires a numerical method for its solution, Newton-Raphson being favourite.

A simpler possibility, but having a slower rate of convergence than Newton-Raphson, is to write the equation as

from which

Now choose a starter value for x, substitute that into the rhs to calculate a new value for x and repeat that until you have the required degree of accuracy.

Starting with x = 2, the sequence is 1.4142, 1.6325, 1.5290, 1.5736, 1.5535, 1.5624,...

but you could use a different starting value.

If x is real root, what is 'a' in the equation

(2+i)x^2+((-97 + 49i)/5-1)x+4-2i=0

$$\\(2+i)x^2+(\frac{(-97 + 49i)}{5}-1)x+4-2i=0\\\\
2x^2+ix^2-\frac{97x}{5}+\frac{49xi}{5}-x+4-2i=0\\\\
2x^2-x+4-\frac{97x}{5}+\frac{49xi}{5}-2i+ix^2=0\\\\
10x^2-102x+20+\left(5x^2+49x-10\right)i=0\\\\$$

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{102}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{10}}\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{5}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{10}}\\
{\mathtt{x}} = {\mathtt{0.2}}\\
\end{array} \right\}$$

$${\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{49}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{5}}}}\\
{\mathtt{x}} = -{\mathtt{10}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{0.2}}\\
{\mathtt{x}} = -{\mathtt{10}}\\
\end{array} \right\}$$

There does not seem to be a single common x value there Alan.

Am I thinking about it wrong, or did I make an error.  (I have not checked my working - I suppose I should.)

EDITED

Thanks Alan - Now it works :)

x=0.2       a = (-97 + 49i)/5

Half of it was pasted half of it was written anyonamous 

No problem took me some time to find the question as my phone was out

Thanks Maia,

I believe in this case x can't be a whole number which means you will have to go through some trial and errors.

when x is 1.5, x^x=1.8371173070873836.... which is obviously smaller than 2

when x=1.6 x^x=2.1212505710975916.... which is bigger than 2

now we know that x is between 1.5 and 1.6

you can always write "x is bigger than 1.5 but smaller than 1.6" or you can do some trial and errors to get more accurate answer such as x=1.559610.....

but i believe you won't be able to get an accurate answer as it is an infinite decimal. 

hope it helped :)

You could also try a = (-97 + 49i)/5

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You will need a numerical approach.  For example:

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$${{\mathtt{x}}}^{{\mathtt{x}}} = {\mathtt{2}}$$

Here is a graph of y=tan(theta)

It illustrates what Geno is telling you  :)