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Hi ProMagma.

I remember we each posted something about offtopic red panda questions that last time you logged in. 

It is not a question, though.

I hope this helps!

Hello Jeff.

Hello Hello Hello.

But (at least right now when I'm typing this) you're question isn't edited!

Oh, so we can trust your question? 

This requires a numerical solution method.  Draw a graph first to see where the solution(s) might lie.

.

Hi Jeff :)

continued here:

17,285

How to solve this equation?

\( {3^{2}}^x+3^x-2=0\)

\({3^{2}}^x+3^x-2=0\\ 3^{(2^x)}+3^x=2\\ 3^{(2^x)}=2-3^x \)

\(2^x\times ln\ 3=ln\ (2-3^x)\)

?

Excuse me!

Hey! Welcome to the forum! 

im new so 300 is a lot to me.

Hi 

If x^2 - 4  is a factor.....then  (x - 2)  and (x + 2)  are  factors

Which means that   2  and - 2 are roots

So

4(2)^4+7(2)^3 +a(2)^2 +b(2) +8  = 0

4(-2)^4+7(-2)^3 +a(-2)^2 + b(-2) +8  = 0

Add these and we have that

128   +  8a   +   16  = 0    ⇒  8a  =  -144  ⇒   a  =  - 18

Subtract them and we have that

112 + 4b   =   0

28 + b  =  0    ⇒   b  = -28 

So....the polynomial is

4x^2  + 7x^3  - 18x^2 - 28x  + 8       and we can write this as

[4x^2  + 7x^3  - 2x^2]  -  16x^2  - 28x + 8  

x^2 (4x^2  + 7x - 2)  -  4 ( 4x^2 + 7x - 2 )

(x^2 - 4) (4x^2 + 7x - 2)

(x^2 - 4) (4x - 1) (x + 2) 

(x - 2) (x + 2) (4x - 1) (x + 2)  =  0

The  values that make this  = 0   are     x= 2, x = 1/4  and x  = -2

If x - 2 is a factor....then  2  is a zero

Thus  

2(2)^3  - a(2)^2  + 5b(2) + 4b  = 0

16 - 4a + 10b  + 4b  = 0

-4a + 14b  = -16

2a - 7b  = 8                   (1) 

And

2(-1)^3  - a(-1)^2  + 5b(-1) + 4b  = -15

-2   - a   - 5b + 4b   = -15

-a - b  =   -13   ⇒   -2a - 2b  =  -26      (2)

Add (1)  and (2)    and we have

-9b  =  -18    ⇒    b  = 2

So

-a - 2  = -13

-a  =  -11

a  = 11

So....the polynomial is

2x^3 - 11x^2 + 5(2)x + 4(2)

2x^3 - 11x^2 + 10x + 8

And we can write

2x^3  [ - 4x^2  -  7x^2 ] + 10x + 8

[2x^3 - 4x^2]  - [7x^2 - 10x - 8 ]

2x^2 (x  - 2)  -  [ 7x^2 - 10x - 8 ]

2x^2 (x - 2) - [ ( 7x  + 4) (x - 2) ]

(x - 2) (2x^2 - 7x - 4)

(x - 2) (2x + 1) ( x - 4)

Sorry.....I believe that exactly 2 should be   6 + 6 +  12   =   24 

a + 980   = 2000    subtract  980  from both sides

a  =  2000 - 980   =  1020

We have the following system of equations

 a    +  b   +   c   =    13    (1)

-a     - b    + c     =    -1     (2)

-8a  -2b    + c     =   -32   (3)

Adding  (1)  and (2) we have that  2c  = 12  ⇒   c  = 6

So (2) and (3) become

-a     -  b   =   -7   ⇒    -2a   -   2b    =    - 14     ⇒  2a  +   2b     =  14      (4)

-8a  -2b   =  -38      (5)

Add (4)  and (5)   and we have that

-6a   =   -24    ⇒   a  = 4

So

a  +  b  + c  =  13

4  +  b   + 6   = 13

b =  3

So    a  =  4, b = 3   and   =  6

helperid1839321, I decided to delve deeper into this subject. You may like my findings!

First of all, I think I understand why this property occurs. I happen to know that \(142857\) is the first 6 digits of the decimal expansion of \(\frac{1}{7}\). At first, I thought this was insignificant, but it turns out that this fact can be used to understand this further. Look at the table below.

This forced me to wonder if there is any more solutions for \(\frac{1}{p}\), where p is a whole number that creates this special property. This is because if another number p causes some repetition, then we would have found another number! YAY!

I decided to enlist some help from a computer here. This is what the computer outputted. 

7, 17, 19, 23, 29, 47

WHAT! There are more! Yes, these have the same property. Let's check them out, shall we?

It appears as if 142857 is the only number that does not start with a zero. Let's keep running this simulation! Thankfully, more numbers output! I let it run for some time, too. 

59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193...

In case you are wondering,

 \(\frac{1}{193}\)=005181347150259067357512953367875647668393782383419689119170984455958549222797927461139896373056994818652849740932642487046632124352331606217616580310880829015544041450777202072538860103626943...

There are a few patterns that I see here 

1) \(p\) must be a prime number

2) The decimal expansion must have a maximum period decimal expansion of \(p-1\).

I am also making a conjecture here that I do not know whether or not is true: there are an infinite number for p that create these types of numbers!

Its great to see you too 

2002 revenue   =  d (1 + x% )   where x is expressed as a decimal

Also OBT,

We usually mark questions as answered as soon as we, or another member, answer them.

If you are not happy with the answer then please just politely say so (along with a thank you for the answerer's efforts) and ask us, a moderator, to remove the tick. We will be quite happy to do this for you.  

Good to see you, PM....welcome back.......!!!

Hi Chris,

I saw you last answer too and I have only just worked out what the question even meant :)

This question is making me think about tthings that I do not think I have considered before.

I might put my 2cents worth in later.

OBT do you undertand the link between Pascal's triangle and combination mathematics?

(CPhill used the combination maths notation in his previous answer )

also OBT,

you should have referenced your last question thread so that people could jump between the two.

Here is th link:

The triangle sum theorem states that the sum of the measures of the interior angles of a triangle equals 180 degrees. Using this theorem alone, one can find the measure of all the angles.

Now, plug in this value for the angle measure expressions.

\(m\angle A=x-16=98-16=82^{\circ}\)

\(m\angle B=2x-155=2*98-155=196-155=41^{\circ}\)

\(m\angle C=\frac{1}{2}x+8=\frac{1}{2}*98+8=49+8=57^{\circ}\)