yeah.. I am not seeing how to work it out with the quadratic formula. Can you elaborate more?
I am a Mathematician
>be faceless void >mfw I have no face
Quadratic equation works once you have your equation in the form ax^2+bx+c. So for this problem, a = 2^x, and b = -5, and c = 1. Then you put those into the quadratic equation and the anrwer is what x equals.
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
?????
Actually in this case, a was 1 and x itself was 2^x.
None.
Didn't there used to be a rule against posting topics for people to do your homework?
None.
No, homework solving threads have been quite common over the past few iterations of SEN.
Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
I remember browsing through some topics in serious discussion, I saw a couple of calculations at your hands. I was hoping you would come along!
I'm only here because they patched SC1 and made it free.
Didn't there used to be a rule against posting topics for people to do your homework?
I wouldn't mind helping to solve homework as long as they ask how and not what is the answer.
Still SEN might not be the right place to ask this kind of questions.
I remember browsing through some topics in serious discussion, I saw a couple of calculations at your hands. I was hoping you would come along!
That's what we math majors are for
?????
Look on first page >_< I want some credits too.
*rawrs
None.
You did help as well, you did basically the same thing as clokr_, he just substituted. His method just clicked a little easier. Thanks for the contribution anyway!
I'm only here because they patched SC1 and made it free.
I always thought factoring was 10x easier than the quadratic
None.
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
I rig my problems so I can factor.
None.
your answer is wrong
. I respect you as a forum member, but that is not the correct way to solve the problem graphically.
Woops, guess I plugged it into WinCalc wrong
(although admittedly I should've been able to easily do that in my head..) All this logarithm and fractional exponents and such were done in 5/6 math last year, but I forgot it
Although I still don't understand why my way of solving doesn't work properly, it seems like it should get the answer..
None.
You can't go from that to this:
At the very least, it would turn into log
2(5-2x) - log
2(2x)
I figured log(1/n) = -log(n), so why not?
None.
Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
I cannot believe I didn't think of solving via substitution.
None.
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
I rig my problems so I can factor.
You have me curious: what's x² + 21x + 85 = 0?