solving unknown exponents

[Vastor]

the question is,

 

Solve the equation [math]2^{x+2} - 2^{x-1} = 28[/math]

 

so, I do the calculation...

 

[math]2^{x+2} - 2^{x-1} = 28[/math]

 

[math]2^{x+2+1} - 2^{x-1+1} = 28 * 2[/math]

 

[math]2^{x+3} - 2^x = 56[/math]

 

[math]2^{x+3} = 2^x + 56[/math]

 

[math]2^x * 8 = 2^x + 56[/math]

 

[math]2^x = \frac{2^x + 56}{8}[/math]

 

[math]2^x = \frac{2^x}{8} + 7[/math]

 

[math]2^x = 2^{x-3} + 7[/math]

 

now i'm stucked, i solve it, but does it's can only be done by wild guess, like this..

 

let x = 3

 

[math]2^3 = 2^{3-3} + 7[/math]

 

[math]2^3 = 2^0 + 7[/math]

 

[math]2^3 = 8[/math]

 

x = 3 #

 

or there any alternative calculation for this?

[mooeypoo]

the question is,

 

Solve the equation [math]2^{x+2} - 2^{x-1} = 28[/math]

 

so, I do the calculation...

 

[math]2^{x+2} - 2^{x-1} = 28[/math]

 

 

Look at what you have here. What if I wrote something like,

 

[math]x^2 + x^4 = 28[/math]

 

What would you do then? Possibly factor out.... something? The 2 power-something in your initial exercise is the same point. Try starting with factoring it.... see if you manage from there. If not, come back and we can continue...

 

TIP: it might help you rewrite the powers in fractions. [math]2^{A-B} = \frac{2^A}{2^B}[/math] ... use it.

[Vastor]

[math]x^2 + x^4 = 28[/math]

 

[math]\frac{x^2}{x^2} + \frac{x^4}{x^2} = \frac{28}{x^2}[/math]

 

[math] 1 + x^2 = \frac{28}{x^2}[/math]

 

 

[math] x^2 = \frac{28}{x^2} -1[/math]

 

[math] \sqrt{x^2} = \sqrt{\frac{28}{x^2} -1}[/math]

 

[math] x = \sqrt{\frac{28}{x^2} -1} [/math]

 

soo... what? i'm not get it...

[timo]

There is no big trick and essentially you were pretty close to the solution already, but then went around and rearranged back to where you came from. In your line [math]2^x * 8 = 2^x + 56[/math] try substracting [math]2^x[/math] on both sides, instead of dividing both sides by 8. It's not the most elegant way, but it is at least your solution for the most part, then.

[mooeypoo]

[math]x^2 + x^4 = 28[/math]

 

[math]\frac{x^2}{x^2} + \frac{x^4}{x^2} = \frac{28}{x^2}[/math]

 

[math] 1 + x^2 = \frac{28}{x^2}[/math]

 

 

[math] x^2 = \frac{28}{x^2} -1[/math]

 

[math] \sqrt{x^2} = \sqrt{\frac{28}{x^2} -1}[/math]

 

[math] x = \sqrt{\frac{28}{x^2} -1} [/math]

 

soo... what? i'm not get it...

 

I didnt say "Divide" -- I said *factor*.

 

 

 

[math]2^{x+2}+2^{x-1}[/math]

 

If I factor out 2^x, I get

 

[math]2^{x} ( 2^2+2^{-1} )[/math]

 

You see where I'm going with this now?

 

Look, you're doing things right, but since you're splitting up the 2^x, you run into problems. Keep the 'x' in one side of the equation and anything WITHOUT an x at the other.

 

That's why I ask you to factor. Try it, I think you're going to see how to do it then -- you're very close.

[Vastor]

thnx all, i got it..

 

Keep the 'x' in one side

 

this give much of the understanding. however, how about the equation you give?

 

[math]x^2 + x^4 = 28[/math] can't keep 'x' in one term, hard to adjust it.

 

 

anyway, i need to improve my algebra skill especially in distributive law,

e.g. before this I even thought that distribute 2x(5 + 2) and 2x(5 * 2) is same

[mooeypoo]

thnx all, i got it..

 

 

 

this give much of the understanding. however, how about the equation you give?

 

[math]x^2 + x^4 = 28[/math] can't keep 'x' in one term, hard to adjust it.

 

 

anyway, i need to improve my algebra skill especially in distributive law,

e.g. before this I even thought that distribute 2x(5 + 2) and 2x(5 * 2) is same

 

Yeah, I gave you this example to demonstrate that you're supposed to leave the x on one side; it wasn't really meant as a full-on drill (though it COULD be solved on a slightly different method).

 

In any case, YOUR equation, just to make it a clear summary, would go something like this:

 

[math]2^{x+2}-2^{x-1}=28[/math]

 

[math]2^{x}(2^{2}-2^{-1})=28[/math]

 

 

[math]2^{x}=\frac{28}{(2^{2}-2^{-1})}[/math]

 

 

 

etc etc, 'till you reach a solution (you did something similar in your first post.)

 

As for the example I gave, the method for a full-on solution is much different. This time it won't help you to put x's and numbers on opposite sides of the equation, you'll likely need to put them all on the same side =0 and look for the quadratic equation format.

 

Sorry for the confusion with my example, I just meant it as a localized "ah-ha" moment for that particular case.. it might've been the wrong example, though

 

~mooey