Science Forum - Very simple but still...

The Science Forum - Scientific Discussion and Debate

Live Chat

FAQ

Usergroups

Science Forum Forum Index » Mathematics » Very simple but still...

Very simple but still...

« View previous topic :: View next topic »

Author

Message

IAlexN

Posted: Fri Mar 05, 2010 10:43 am Post subject: Very simple but still...

Forum Freshman

Joined: 24 Nov 2009
Posts: 77

Hi, I'm really struggling with this question.

2^(square root of x) = 8

I know that x, just by quickly looking at the equation is, is 9. However, I've no idea how I can solve this equation; showing that x = 9.

Could someone please help me.

DrRocket

Posted: Fri Mar 05, 2010 10:59 am Post subject: Re: Very simple but still...

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

IAlexN wrote:

$2^{\sqrt x} = 8$

$log_2(2^{\sqrt x}) = \sqrt x = log_2(<img src="images/smiles/icon_cool.gif" alt="Cool" border="0" /> = log_2(2^3) = 3$

$x = (\sqrt x)^2 = 3^2 = 9$

IAlexN

Posted: Fri Mar 05, 2010 11:06 am Post subject:

Forum Freshman

Joined: 24 Nov 2009
Posts: 77

Thank you for you reply DrRocket. However is there any other way to solve the equation, without using logarithm. Because, I haven't started to study logarithm yet.

Heinsbergrelatz

Posted: Fri Mar 05, 2010 11:11 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

this is the approach my maths teacher tells the ordinary or the standard mathematics level students in our year level.

$2^{\sqrt{x}} = 8$

$2^{x^{\frac{1}{2}}} = 2^{3}$

same base thereby;

$x^{\frac{1}{2}} = 3$

square to both sides to eliminate the $\frac{1}{2}$ ;

$x^{({\frac{1}{2}})^{2}} = 3^{2}$

x=9

i cannot think of any other way than this.
_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

DrRocket

Posted: Fri Mar 05, 2010 11:14 am Post subject:

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

Heinsbergrelatz wrote:

this is the approach my maths teacher tells the ordinary or the standard mathematics level students in our year level.

$2^{\sqrt{x}} = 8$

$2^{x^{\frac{1}{2}}} = 2^{3}$

same base therby;

$x^{\frac{1}{2}} = 3$

square to both sides to eliminate the $\frac{1}{2}$ ;

$x^{{\frac{1}{2}}^{2}} = 3^{2}$

x=9

i cannot think of any other way than this.

That works. You have implicitly used the logarithm base 2, but that is perfectly OK.

IAlexN

Posted: Fri Mar 05, 2010 11:30 am Post subject:

Forum Freshman

Joined: 24 Nov 2009
Posts: 77

Thanks for your answer Heinsbergrelatz, it was very helpful Very Happy

DrRocket wrote:

That works. You have implicitly used the logarithm base 2, but that is perfectly OK.

However, just out of curiosity what does logarithm base 2 actually mean?

Last edited by IAlexN on Fri Mar 05, 2010 11:32 am; edited 1 time in total

Heinsbergrelatz

Posted: Fri Mar 05, 2010 11:32 am Post subject:

Forum Masters Degree

Joined: 01 Aug 2009
Posts: 598
Location: Singapore

Quote:

Thanks for your answer Heinsbergrelatz, it was very helpful

Your welcome Very Happy

_________________
------------------

$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

Arcane_Mathematician

Posted: Fri Mar 05, 2010 1:14 pm Post subject:

Forum Cosmic Wizard

Joined: 21 Feb 2009
Posts: 2478
Location: San Jose, California

IAlexN wrote:

Thanks for your answer Heinsbergrelatz, it was very helpful Very Happy

DrRocket wrote:

That works. You have implicitly used the logarithm base 2, but that is perfectly OK.

However, just out of curiosity what does logarithm base 2 actually mean?

is a way to rewrite a^x=b

. Logarithm base 2 just sets up a function to determine what 2^x=y

is at some y. It isolates x in the equation in a solvable way.
_________________
I post here to spur conversation, instead of just looking up what I want to know

Plato wrote:

Wise men talk because they have something to say; fools, because they have to say something.

Idiocracy wrote:

Welcome to Costco, I love you

DrRocket

Posted: Fri Mar 05, 2010 2:28 pm Post subject:

Forum Radioactive Isotope

Joined: 12 Aug 2008
Posts: 3114

Arcane_Mathematician wrote:

IAlexN wrote:

Thanks for your answer Heinsbergrelatz, it was very helpful Very Happy

DrRocket wrote:

That works. You have implicitly used the logarithm base 2, but that is perfectly OK.

However, just out of curiosity what does logarithm base 2 actually mean?

is a way to rewrite a^x=b

. Logarithm base 2 just sets up a function to determine what 2^x=y

is at some y. It isolates x in the equation in a solvable way.

precisely

Ellatha

Posted: Fri Mar 05, 2010 7:49 pm Post subject:

Forum Junior

Joined: 11 Jul 2009
Posts: 210

IAlexN wrote:

Thanks for your answer Heinsbergrelatz, it was very helpful Very Happy

DrRocket wrote:

That works. You have implicitly used the logarithm base 2, but that is perfectly OK.

However, just out of curiosity what does logarithm base 2 actually mean?

Have you studied growth and decay yet?

Display posts from previous:

Page 1 of 1

Science Forum Forum Index » Mathematics » Very simple but still...

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum