easy problem

[Safari]

How do I solve this one?

5^2x - 5^2 = 5^3

This is what I've done

5^2x - 5^2 = 5^3
5^2x = 150

Then I thionk i should use ln but not sure how.

Thnx

[Alvaro]

5^2x = 150

Then I thionk i should use ln but not sure how.

That's right. Take log on both sides and you get

log(5^2x)=log(150)

But you should know that log(a^b)=b*log(a), so

2x*log(5)=log(150)

x = log(150)/(2*log(5)) = 1.556641376 approx.

[Safari]

Thank you

[Safari]

Is it okay if I post my simple question here or is it for more advance posts so that everyone can try the problem?

I've got another easy problem (easy for you).

x^4 - 3x^2 + 2 = 0

what I've done:

y = x^2 =>

=> y^2 - 3y + 2

(3/2) +- sqrt((3/2)^2 - 2)
=
1,5 +- sqrt(0,25)
=
1,5 +- 0,5

y1 = 2, y2 = 1

=>(x^4 - 3x^2 + 2 = 0, y = x^2) =>

=> x1 = 2^2 - 3*2 + 2 = 0
=> x2 = 1^2 - 3*1 + 2 = 0

But thats impossible. Both can't be 0.

Thanx in advance

[Alvaro]

Is it okay if I post my simple question here or is it for more advance posts so that everyone can try the problem?

I've got another easy problem (easy for you).

x^4 - 3x^2 + 2 = 0

what I've done:

y = x^2 =>

=> y^2 - 3y + 2

(3/2) +- sqrt((3/2)^2 - 2)
=
1,5 +- sqrt(0,25)
=
1,5 +- 0,5

y1 = 2, y2 = 1

So far, everything is correct.

=>(x^4 - 3x^2 + 2 = 0, y = x^2) =>

=> x1 = 2^2 - 3*2 + 2 = 0
=> x2 = 1^2 - 3*1 + 2 = 0

I don't know what you are doing there. Once you have the values of y, try to find which values of x make y=1 or y=2. They are +1, -1, +sqrt(2) and -sqrt(2). Those are the solutions to the first equation.

[Safari]

Sorry but you lost me there

Do you mean?

where

x^4 - 3x^2 + 2 = 1
x^4 - 3x^2 + 2 = 2

=>
A)
x^4 - 3x^2 + 1
(x^4)^(1/2) - (3x^2)^(1/2) + 1^(1/2) //Trying to make x^4 => x^2
=>
x^2 - 3x + 1
=>
(3/2) +- sqrt((3/2)^2 - 1)
1,5 +- sqrt(5/4)

But it becomes wrong

[Alvaro]

Sorry but you lost me there

Well, re-read my post 10 times, or until it makes sense (whichever happens first). Then post again.

Do you mean?

where

x^4 - 3x^2 + 2 = 1
x^4 - 3x^2 + 2 = 2

No, I have no clue what you are doing there.

[Safari]

It took 3 times for me to read it before i got that "ahhaaa" feeling
sorry

[Alvaro]

It took 3 times for me to read it before i got that "ahhaaa" feeling
sorry

Hey, it happens to everybody sometimes. I only look upset because of the scary green monster. Let's see if a few smileys make you feel better.

[Safari]

Here I come again (trying to understand every problem in this book:) )

(6/x) - ((x+15)/(x+1)) + 2 = 0

((6(x+1))/(x(x+1))) - (((x(x+15))/(x(x+1))) + ((x(x+1))2)/(1(x(x+1))) = 0 //Makes the bottom x^2 + x on them all

(6x+6)/(x^2+x) - (x^2 + 15x)/(x^2+x) + (2x^2+2x)/(x^2+x) = 0

(6x+6) - (x^2+15x) + (2x^2+2x) = x^2+x

6x+6 - x^2-15x + 2x^2+2x = x^2+x

6x+6 - x^2-15x + 2x^2+2x - x^2-x = 0

-8x + 6 == 6 - 8x

2(3-4x) = > x = 1.333...

But the answer should be 1 and 6

Thanx in advance

[Alvaro]

(6x+6)/(x^2+x) - (x^2 + 15x)/(x^2+x) + (2x^2+2x)/(x^2+x) = 0

(6x+6) - (x^2+15x) + (2x^2+2x) = x^2+x

Where did the part to the right of the `=' come from?

[Safari]

Edit: Yaaa now I got it...Thank you

[OgreCoder]

The problem is that (x^2 +x) * 0 = 0. I think you thought about adding it to the right side.

[Safari]

The problem is that (x^2 +x) * 0 = 0. I think you thought about adding it to the right side.

Yes that was what I first thought

[Safari]

Hi again

I saw this exvation:

5sinx - cosx = 0

And though "I have never seen that before". So I sat down and try to figure it out. So I did like this

5sinx = cosx
5sinx = 1cosx
sinx/cosx = 1/5
tanx = 1/5

And i works! But it's one thing I dont understand. When I move cosx to the right side of '=', it becomes '+'. But when I move it back it becomes '*' why is that? I have always woundering this thing:

a = b

Will it then be:
a*b = 0

or

a+b = 0?

Thanx in advance