seeking a math guru

[MrMageo]

Ok so i am many years removed from highschool and have decided to return to college. For this I need some grade 12 math upgrade (too much mary jane in highschool ). Anyhow math is alot different now then I remembered and I am literally beating my head against the wall on this new topic. If someone can explain it to me that would be great (the teacher is confusing when she explains it and 95% of the class is as lost as I am). It is like quadratics with exponents. Anyhow here is a sample question. (^ = exponent)

(2^2x)-7(2^x)-8=0

now from what I understand I need to find common factors which will equal -8 when multiplied and -7 when added

(-8) x 1 = -8
(-8) + 1 = -7

so to me it looks like I have 1, (-8) as my factors.

Now how do I fit them into this equation ive racked my brain and searched the interwebs and can't find a similar example to help me.

According to the answer sheet X=3 but ive had so many wild answers all except 3.

If someone who inderstands this stuff can explain it to me in steps Id really appreciate it. Thanks in advance.

[Anaki]

Re: seeking a math guru

um i was gonna try and write out an answer but i got stuck. but um since your using exponents you may wanna use the log function. since 2^x is 2*2 a total of x times. as opposed to 2+2 a total of x times which is the math you use to get that -8 and -7
______________________________
(2^2x)-7(2^x)-8=0

(2^2x)-8 = 7(2^x)

((2^2x)-8)/(2^x) = 7

(2^x) - 8/(2^x) = 7
then from there you can inda use substitution to say

a= 2^x.
thus
a- 8/a = 7.
so a^2 -8 = 7a.

a^2 -7a -8=0
so a=8. thus
2^x = 8.

then tae the base 2 log which will be
log (2^x) / log(2) = x
so log(8)/log(2) = 3
and thus x equals 3.


edit: i few things i did there was

(a-b)/c = a/c - b/c

log (or ln both work if your gonna go for base 2 log(a)/log(2) and ln(a)/ln(2) are the same

[Mog]

Re: seeking a math guru

Well, I attempted for a good 5 minutes then gave up. T.T Just out of curiousity, what class is this? I've never had to solve anything that looks like that, and I'm a math major.

This looks like a logarithm problem...but the real problem lies in too many terms. Usually with logs, if you have something like 2^x = 6, you can take the logarithm of both sides:

log(2^x) = log 6
x log 2 = log 6
x = log 6 / log 2

But in this case, that minus sign screws it up. Also, you can't simply turn everything into a like base because there are too many terms. >< There might be some funky algorithm involved that I've long forgotten. Who knows.

[MrMageo]

Re: seeking a math guru

well im going by what I thought i understood. we were shown a thing called MAN

M= Multiple
A= Addition
N= numbers (comon number that multiply to and add to A)

in this question from what I can tell

M= -8
A= -7
N= 1,-8

since we know

M= -8 (1x-8)
A= -7 (-8+1)

Some how it gets broken down into 2 equations with

2^2x(-7) and 2^x(-8)

now i dont know if the 2's cancell each other out or not

putting the equation at ^2x(-7) and ^x(-8)= -4 (-2,-2 to cancel out)

then I would go with cancelling out the -7,-8

^2x^x- 11 (-4 + 7 + 8)
^3x = 11
3x/3 = 11/3

x= 11/3

which dosent work and I didnt use 1, or -8

Im totally lost (I hate this shit)

Apparently its grade 11 review math (we have covered the factoring of simple equations 2x+8x+12, and various other things)

[Anaki]

Re: seeking a math guru

*wonders if they only saw his first post and not his second post (which got automerged) which has the answers *as a note the k key on this keyboard is broken so it gets missed alot.

[Mog]

Re: seeking a math guru

*wonders if they only saw his first post and not his second post (which got automerged) which has the answers *as a note the k key on this keyboard is broken so it gets missed alot.

Your second post wasn't up yet. =p

Ahh, I see what you did. Basically, just moving terms around and getting them into the same form. I knew I was looking at it too hard. T.T

[Anaki]

Re: seeking a math guru

ya first time i did it i went log (a-b) is um CHECK WIKI then CRAP its even more complex. then i gave up :X and then it hit me...

[MrMageo]

Re: seeking a math guru

thanks alot it kinda makes sense to me now i never thought to check the wili either I only use it for ffxi stuff so it slipped my mind

[Nuriko]

Re: seeking a math guru

Ok so i am many years removed from highschool and have decided to return to college. For this I need some grade 12 math upgrade (too much mary jane in highschool ). Anyhow math is alot different now then I remembered and I am literally beating my head against the wall on this new topic. If someone can explain it to me that would be great (the teacher is confusing when she explains it and 95% of the class is as lost as I am). It is like quadratics with exponents. Anyhow here is a sample question. (^ = exponent)

(2^2x)-7(2^x)-8=0

now from what I understand I need to find common factors which will equal -8 when multiplied and -7 when added

(-8) x 1 = -8
(-8) + 1 = -7

so to me it looks like I have 1, (-8) as my factors.

Now how do I fit them into this equation ive racked my brain and searched the interwebs and can't find a similar example to help me.

According to the answer sheet X=3 but ive had so many wild answers all except 3.

If someone who inderstands this stuff can explain it to me in steps Id really appreciate it. Thanks in advance.

Actually, you just sort of flipped a sign.

Let y = 2^x (just to make life easier)

y^2 - 7y - 8 = 0

(y + 1) (y - 8) = 0

y = -1, y = 8

That gives x = 3 for y = 8, and a total mess for y = -1, which I'm fairly sure involves imaginary numbers and exponents and/or logarithms...

[Sabaron]

Re: seeking a math guru

(2^2x)-7(2^x)-8=0

Here we go (this is sans textbook, so I can be wrong).

L2 is the base 2 logarithm

(2^2x)-7(2^x)= 8

L2 ((2^2x)-7(2^x))=L2(8)

L2(2^2x) - L2(7(2^x)) = 3

2x - L2(7)x = 3

(2 - L2(7))x = 3

x = 3/(2-L2(7))

but for that stupid 7, everything is relatively clean. The textbook writers should've just made is 16 or something nice.

[Icemage]

Re: seeking a math guru

Ok so i am many years removed from highschool and have decided to return to college. For this I need some grade 12 math upgrade (too much mary jane in highschool ). Anyhow math is alot different now then I remembered and I am literally beating my head against the wall on this new topic. If someone can explain it to me that would be great (the teacher is confusing when she explains it and 95% of the class is as lost as I am). It is like quadratics with exponents. Anyhow here is a sample question. (^ = exponent)

(2^2x)-7(2^x)-8=0

now from what I understand I need to find common factors which will equal -8 when multiplied and -7 when added

(-8) x 1 = -8
(-8) + 1 = -7

so to me it looks like I have 1, (-8) as my factors.

Now how do I fit them into this equation ive racked my brain and searched the interwebs and can't find a similar example to help me.

According to the answer sheet X=3 but ive had so many wild answers all except 3.

If someone who inderstands this stuff can explain it to me in steps Id really appreciate it. Thanks in advance.

(2^2x)-7(2^x)-8=0

This isn't a logarithm problem. It's just simple algebra.

Let Y = 2^x

(2^x)^2 - 7(2^x) - 8 = 0
Y^2 - 7Y - 8 = 0

Factor the above into:

(Y + 1)(Y - 8) = 0

This produces two possibilities:

Y + 1 = 0
OR
Y - 8 = 0

Thus, Y = -1 or Y = +8

However, Y = -1 is not a valid result. Why? Because there's no way to solve this:

2^x = -1

An exponent will always produce a positive result if the base number is positive. So we throw out Y = -1 as a possible result.

This leaves our answer as Y = 8

Y = 8
2^x = 8

But 8 = 2^3

So...

2^x = 2^3

x = 3

There's your answer.


Icemage

[Nuriko]

Re: seeking a math guru

Incidentally, that other one, where 2^x = -1?

Given e^(pi*i) = -1

e^(pi*i) = 2^x

Since e^ln(y) = y

e^(pi*i) = e^ln(2^x) = e^(x * ln 2)

x * ln 2 = pi * i

x = i*(pi / ln 2)

[Sabaron]

Re: seeking a math guru

(2^2x)-7(2^x)-8=0

This isn't a logarithm problem. It's just simple algebra.

Let Y = 2^x

(2^x)^2 - 7(2^x) - 8 = 0
Y^2 - 7Y - 8 = 0

Factor the above into:

(Y + 1)(Y - 8) = 0

This produces two possibilities:

Y + 1 = 0
OR
Y - 8 = 0

Thus, Y = -1 or Y = +8

However, Y = -1 is not a valid result. Why? Because there's no way to solve this:

2^x = -1

An exponent will always produce a positive result if the base number is positive. So we throw out Y = -1 as a possible result.

This leaves our answer as Y = 8

Y = 8
2^x = 8

But 8 = 2^3

So...

2^x = 2^3

x = 3

There's your answer.


Icemage

(2^(2*3))-7(2^3)-8= 0

Yar... I be wrong. I should've substituted back -.-

[Icemage]

Re: seeking a math guru

e^(pi*i) = -1

Is there a proof for this somewhere? I've never come across that particular definition, and it seems a bit... odd.

At any case, I don't think a remedial math course is going to venture into imaginary number solutions to algebraic problems... :p


Icemage

[Mog]

Re: seeking a math guru

Is there a proof for this somewhere? I've never come across that particular definition, and it seems a bit... odd.

At any case, I don't think a remedial math course is going to venture into imaginary number solutions to algebraic problems... :p


Icemage

There's a bunch of crazy formulas out there involve e, pi, cos, sin and imaginary numbers. This is probably some of the stuff I'll be looking at with 4000 level math.

If you really want to delve into it and try to even understand it, google Leonhard Euler. The proofs are beyond me. <.>