solving for x - Math Help Forum

$Math Help Forum$

$Go Back$		Math Help Forum > Pre-University Math Help > Pre-Calculus
$Reload this Page$ solving for x

Download MHF Flyer

Mark Forums Read

$Reply$

#1

$Old$ November 21st, 2008, 01:12 AM

Kitty216 $Kitty216 is offline$

Junior Member

Join Date: Aug 2008

Posts: 40

Country:

United States

Thanks: 19

Thanked 0 Times in 0 Posts

$Kitty216 is on a distinguished road$

$Default$ solving for x

Solve for x

It looks so simple but when I tried it, I ended up getting strange numbers. Help!

$Reply With Quote$

Advertisement

#2

$Old$ November 21st, 2008, 02:31 AM

$mr fantastic's Avatar$

mr fantastic $mr fantastic is online now$

Flow Master

Join Date: Dec 2007

Location: Zeitgeist

Posts: 12,007

Country:

Australia

Thanks: 2,513

Thanked 4,694 Times in 4,131 Posts

$mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$

$Default$

Quote:

Originally Posted by Kitty216 $View Post$

Solve for x

It looks so simple but when I tried it, I ended up getting strange numbers. Help!

Here's an outline - you can fill in the details.

You have $(2^2)^{2x} - 2 \cdot (2^2)^{x+4} + (2^2)^8 = 0$ .

Substitute $w = 2^x$ and do a bit of simplifying.

You end up with $w^2 - 2^9 w + 2^{16} = 0$ .

This factorises: $(w - 2^8)^2 = 0$ .

Therefore $2^x = 2^8$ .

__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

Last edited by mr fantastic; November 21st, 2008 at 02:45 AM.

$Reply With Quote$

#3

$Old$ November 21st, 2008, 02:42 AM

Kitty216 $Kitty216 is offline$

Junior Member

Join Date: Aug 2008

Posts: 40

Country:

United States

Thanks: 19

Thanked 0 Times in 0 Posts

$Kitty216 is on a distinguished road$

$Default$

I'm kinda confused how you substituted

in and got

$Reply With Quote$

#4

$Old$ November 21st, 2008, 02:48 AM

$mr fantastic's Avatar$

mr fantastic $mr fantastic is online now$

Flow Master

Join Date: Dec 2007

Location: Zeitgeist

Posts: 12,007

Country:

Australia

Thanks: 2,513

Thanked 4,694 Times in 4,131 Posts

$mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$ $mr fantastic has a reputation beyond repute$

$Default$

Quote:

Originally Posted by Kitty216 $View Post$

I'm kinda confused how you substituted

in and got

$(2^2)^{2x} - 2 \cdot (2^2)^{x+4} + (2^2)^8 = 0$

From the usual index laws:

$\Rightarrow (2^{2x})^2 - 2^{2x + 9} + 2^16 = 0$

$\Rightarrow (2^{2x})^2 - 2^9 \cdot 2^{2x} + 2^16 = 0$

Now make the substitution.

__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

$Reply With Quote$

The following users thank mr fantastic for this useful post:
Kitty216
$Donate to MHF$

#5

$Old$ November 21st, 2008, 03:01 AM

Kitty216 $Kitty216 is offline$

Junior Member

Join Date: Aug 2008

Posts: 40

Country:

United States

Thanks: 19

Thanked 0 Times in 0 Posts

$Kitty216 is on a distinguished road$

$Default$

ohh okay, now I got

how do I factor that?

doesn't seem to work.

ahhh.. nvm! I got it now! It does work! =D Thanks so much!!

Last edited by Kitty216; November 21st, 2008 at 03:06 AM. Reason: got the answer

$Reply With Quote$

$Reply$

« Polynomial function | Top | equation of perpendicular line »

Thread Tools
$Show Printable Version$ Show Printable Version $Email this Page$ Email this Page
Display Modes
$Linear Mode$ Linear Mode $Hybrid Mode$ Switch to Hybrid Mode $Threaded Mode$ Switch to Threaded Mode

Posting Rules
You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are Off

Forum Jump

All times are GMT -7. The time now is 09:46 PM.

your link here Sponsors

rv solar, e-learning,

Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.