The problem: Solve algebraically, support your answer graphically.

2^x + 2^(-x) = 5

Well, so far i have solved it graphically. The answer is ≈±2.260
But for the life of me, I cannot solve this algebraically. Please help someone, and show your steps. thanks.

(Also, i have another one but this one isn't as important)
y = (x+3) / (x-2)
I cannot solve it either

It's easier to solve the bottom one graphically as it is now, just replace x with numbers until you have a graph. Assuming we are solving for x,
Y=(x+3)/(x-2) then multiply both sides by x-2
Xy-2y=x+3
Then subtract x from both sides, and add 2y to both sides
xy-x=3+2y seperatd the x
X(y-1)=3+2y then divide both sides by y-1
x=(3+2y)/(y-1)

For the top one, i'm not sure. I'll post when i've got it.

I appreciate the help zany. Now that you show it to me it is kind of silly that i couldn't solve it. Haha, i got that far but it didn't occur to me to add 2y to both sides when i needed to.

No problem. For the top one, I haven't done that sort before, have you got anywhere with getting the x's down?

This book has odd answers in the back. This is problem # 36. I looked at problem # 35, it is e^x+e^-x=3. That is awfully similar to problem # 36.

The solution to problem #35 in the back of the book is x = ln ( 3 ± sqrt(5) / 2). I honestly have no clue how this got here.

I am pretty sure that I am supposed to use logs somehow. I remember my teacher telling us that whenever you have a=b^x, you must use a log in this manner:

Log(base b) of a = x

Our calculators work in base 10, so unless b=10, you can use change of base to solve the rest using the calculator:

(log a)/(log b) = x

The problem solving it this way: X is in the exponent spot twice.

So here is my question, what is the log correlation between a=b^x and a=b^x + b^-x

could it be.... log(base b) a + log(base b) a = ±X

I THINK ITS IT, LET ME SOLVE BRB.

Yeag Ok i see it now, i gotta go out soon but i'll give it a go.
Incidentally is it a saxon maths book?

no, but i think all math books have odd answers in back.

Random. I'm not allowed to look back there though xD. Did it work?

no

Assuming that this homework isn't being graded based on correctness, just try to do it and if you can't get it then try to get it explained in class.

Our teacher operates on very different grounds. You are correct, it is not graded based on correctness, but that doesn't mean that it will be explained in class. She (my teacher) wants students to explain to students. She does very little of this. I am not prideful when I say that I am the most capable person in my class of 9 people. If i don't figure it out this weekend, I probably will end up with a very hazy understanding. I will be tenacious with this problem until my fingers bleed.

2^1 + 2^-1 = The X Root of 5
2 + .25 = The X Root of 5
2.25 = The X Root of 5
Plug the X Root of 5 into a graph, where the graph hits y = 2.25, that is X.

I swear I learned how to do this last year, but I forgot how to solve that end part, so I cheated

You're good, I was looking at that forever and I still didn't know what to do, stupid logs. lol.

Going back to problem #35. the origional problem is 3=e^x+e^(-x).
The answer is:

x= ln((3±√5)/2)

In order to solve 5=2^x+2^-x, i could try to reverse engineer the answer for 35 and then re-use the steps on 36. Here is the process of reversing the answer.

(a=b^x) = (log(base b) a = x)

Therefore in problem #35:
a = (3±√5)/2)
b = e
so...

(3±√5)/2) = e^x (multiply both sides by two)
(3±√5) = e^x + e^x (okay, looks semi-familiar... Now what?)

The thing is, i have no clue how the book got to their answer. Where did √5 come from? and the ±?

Take the natural log of both sides and you have your answer? lol. I'm not too sure, what kind of Algebra is this, high school?

Wtf how did I get to cosh?!?!

Well, its the "review" section of AP calc. and yeah, high school.

@Falkoner your answer is wrong . I respect you as a forum member, but that is not the correct way to solve the problem graphically.


2^-1 = .5 not .25

And hell, even if you took the x root of 5 and find where the graph hits 2.5 its still incorrect. y1 = 2^x + 2^-x, y2 = 5, calculate intersect. That's how you find the answer correctly

Post has been edited 1 time(s), last time on Aug 30 2009, 4:41 am by ProtoTank. Reason: damn typos

I got a different answer than you for the first one algebraically. But I probably messed it up somewhere.

2^x +2^-x = 5

2⁰ + 2^-2x = 5/2^x

2^-2x = 5-2^x/2^x

log₂(5-2^x/2^x) = -2x

-(5-2^x)log₂(2^x) = -2x

-5x + (x)(2^x) = -2x

2^x = 3

I think the flaw lies in me using the reciprocal rule for logarithms wrong. Oh well, DTBK will be along soon enough to solve it.

I actually tried to solve it a bunch of different ways, with various obscure rules, and always ended back at the beginning so it might not be solvable without brute forcing.

You can't go from that to this:
At the very least, it would turn into log₂(5-2x) - log₂(2x)

I shall attempt a failure.
2^x + 2^(-x) = 5
multiply both ends by 2^x
2^2x + 1 = 5*2^x
2^2x - 5*2^x + 1 = 0
subtitute 2^x with a big X ( X = 2^x )
X^2 - 5X + 1 = 0
I don't know what you call this in English, because I am not a native English speaker but
(5±21^[1/2])/2 = X
2^x = (5±21^[1/2])/2
log(2^x) = log(5±21^[1/2])/2
xlog2 = log(5±21^[1/2])/2
x = (log(5±21^[1/2])/2)/log2
x = log₂(5±21^[1/2])/2
x = log₂(5±21^[1/2]) - 1

I think thats it.

Oh yeah. I think I shall attempt another failure, not that I have taken notice.
(3±√5) = 2e^x
(3±√5)/2 = e^x
ln((3±√5)/2) = x ( since an ln is log(base e), a ln(e^x) should be x. I think your #36 is a tad different than this one, while in principal- the same. )

Post has been edited 1 time(s), last time on Aug 30 2009, 7:42 am by BeDazed.