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/sci/ - Science & Math

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4110080 No.4110080 [Reply] [Original]

How do I solve this equation?
2^(2x+1)-2^(x+2)+8=0

I got -2. Wolfram says no solution.
It's on the practice so clearly I'm putting it into wolfram wrong or something.

>> No.4110100

or is that the answer? no solution?

>> No.4110108

>>4110080
There are complex solutions

>> No.4110113

>>4110108
this is pre calc, we don't consider complex solutions

>> No.4110120

there seems to be no solution.

Try substituting in u = log(base 2)x

Then you get a quadratic you can solve for u and then x but I think you get

2u^2 - 4u + 8 = 0

but that has no solution

>> No.4110121
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4110121

>>4110113
>we don't consider complex solutions

>> No.4110128

Stupid niggers, change 8 into 2^3 and drop all of the bases setting the exponents into an equation.

>> No.4110141

>>4110128
I did that and got -2, but wolfram says that's wrong.

>> No.4110147

>>4110128
what the fuck do you mean drop the bases you fucking moron

>> No.4110157

>>4110147
he thinks he can take log base 2 of everything, and log base 2 of 2^x = x, but that doesn't work, because log base 2 doesn't distribute nicely like that.

>> No.4110169

>>4110141
It's obviously not -2

>> No.4110182

...did you plug in -2 to the original equation?

>> No.4110179

>>4110169
yeah, I tried working it out on my own, substituting -2 for the x's. it's obviously not -2.

>> No.4110178

<span class="math">2^(2x+1)-2^(x+2)+8=0[/spoiler]
Divide by 8
<span class="math">2^(2x-2)-2^(x-1)+1=0[/spoiler]
Factor
<span class="math">(2^(x-1))^2-2^(x-1)+1=0[/spoiler]
Should I go farther or you got it now?

>> No.4110192

>>4110182
yes, and found it didn't work.

>> No.4110196

>>4110178
Sorry, fucked up the curly braces
<span class="math">2^{(2x+1)}-2^{(x+2)}+8=0[/spoiler]
Divide by 8
<span class="math">2^{(2x-2)}-2^{(x-1)}+1=0[/spoiler]
Factor
<span class="math">2^{(x-1)}^2-2^{(x-1)}+1=0[/spoiler]
Should I go farther or you got it now?

>> No.4110212

>>4110196
phocc, last one should be
<span class="math">(2^{(x-1)})^2-2^{(x-1)}+1=0[/spoiler]

>> No.4110226

No real solutions. It was a typo or troll.

>> No.4110265

>>4110226
well the prof trolled me good. the pic there is right from the practice exam. :/

>> No.4110298

>>4110265
http://www.wolframalpha.com/input/?i=2%5E%282x%2B1%29+-+2%5E%28x%2B2%29+%2B+8+%3D+0

>> No.4110302

What makes the empty set any less satisfactory as a solution?

>> No.4110310

2^(2x+1)-2^(x+2)+8=0
2^(2x+1)-2^(x+2)=-(2)^3
2x+1-(x+2)=-3
x-1=-3
x=-2

>> No.4110309

>>4110265
No real solutions is a proper answer.

>> No.4110326

>>4110310
@step 3, not sure if troll or very stupid

>> No.4110332

>>4110265
What was the actual question? All we see is something that has a number and an equation. Does it say "solve for x"?

>> No.4110343

Holy shit you guys are dumb.

you can change the equation to be:

2^(2x+1) + 2^3 = 2^(x+2)
then you can see that as a property of exponents: 2x + 1 + 3 = x + 2

Then the solution is elementary

>> No.4110349
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4110349

>>4110343

>> No.4110354

OP, have you considered using logarithms?

>> No.4110364 [DELETED] 
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4110364

Wow, this thread is really exposing the idiots of /sci/.

NO REAL SOLUTIONS.

>> No.4110386

You do know
that in math sometimes, DNE or
does not exist
is a valid answer or solution to a problem

As shown by the dude who posted the wolframalpha link
http://www.wolframalpha.com/input/?i=2%5E%282x%2B1%29+-+2%5E%28x%2B2%29+%2B+8+%3D+0
Its quite obvious it NEVER crosses 0
if it doesnt cross 0, then it doesnt ever equal 0?


What you need to show us
is the actual question to the problem
Does it actually say "solve for x"
or does it say "solve"
or does it say prove? or disprove? need more information!

>> No.4110392

>>4110354
I went ahead and used Logarithms to see for myself anyway. This is what I got: http://mathbin.net/84464.. I've probably made a mistake though as it's 6:33am, and I got a couple hours sleep.

>> No.4110395

>>4110343
Yes because 2^1 + 2^2 = 2^3, or 2+4=8

>> No.4110413

2^(2x+1) - 2^(x+2) + 8 roots
http://www.wolframalpha.com/input/?i=2%5E%282x%2B1%29+-+2%5E%28x%2B2%29+%2B+8++roots

it gives you what x =

decimal form
http://www.wolframalpha.com/input/?i=2%5E%282x%2B1%29+-+2%5E%28x%2B2%29+%2B+8++roots+complex

x approximately equals = 1.4427i *6.28319n - (1.0472+0.693147i)
or
x = 1.4427i *6.28319n + 1.0472-0.693147i

Dur durr solved!
also using WolframAlpha
lrn2use wolfram properly

>> No.4110428

>>4110395
2^1 + 2^2 = 2^3
but
1+2 = 3 does work!
you did it differently then your ref'd post

>> No.4110503

speaking of logs, what is log(log(x) simplified?

>> No.4110588

>>4110332
give solutions for x

>> No.4110597

>>4110503
assuming you mean log base 10 of log base 10 of x,
log base 100 of x?

>> No.4110734
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4110734

>>4110080
Hey OP, check'em http://rechneronline.de/function-graphs/
the equation has no solution

>> No.4110735
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4110735

>>4110734
in reals