2^(x+2) + 3^x = 320...

My graphing calc says 4.845... I just want to know how they figure the thing out on paper... please help a fellow board member...

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transientb.bandcamp.com

Okay I figured it had something to do with taking the natural log of both sides and dropping the exponents, but my head hurts and I can't get it to work. I keep getting approximately 2.44 every time I do it, which when checked out is not even close to correct. Sorry

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ok if you can tell me how to solve

a^1.58 + 4a = 320

then I will be able to solve the bastard once and for all

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transientb.bandcamp.com

Well im quite stumped, i got the same answer for your first post doyle. But when i went on to paper for your second post i gave up after i got to (a^1.58)/4 + a =80 :/

Ile post a proper solution tomorow after i figure it out.

Ok lets start at something easier....

2^x + 3^x = 13

Guess and check shows the answer is x = 2, but there must be an actual formula to follow...

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transientb.bandcamp.com

Since you've reduced it to

a^1.58 + 4a - 320 = 0

Use Newton's method ( x2 = x1 - f(x1)/f'(x1) ) until your happy with the precision of your answer.

I'm not sure if it can be done algebraically, but my gosh that's a long time ago for me

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You guys are all way too smart for me.


Math is hard!

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I love the way u americans say math, its so quaint! Turns out i had a math assesment test today, i failed it miserably so ile pick up a Lovely U some time next week *smiles*

I'm on the most far end of the spectrum of verbal vs math as it gets. Words > Numbers.

Where is DDRkirby when we need him?

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huh, I only learned how to use Newton's Method only in instances where the limit of an equation at f(x) was 0/0. I hate when math classes only teach you shortcuts...

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if we could find the F-1(x) we'd have no reason to even use newtons law...

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transientb.bandcamp.com

Ummm, I'm not sure about your terminology, but that really sounds kind of limiting. Does that mean where the x-axis is a tangent to some local min/max? Anyway that's why I rearranged it so that the 320 was on the LHS of the equation, hence

f(a) = a^1.58 + 4a - 320
and
f'(a) = 1.58a^0.58 + 4

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I have no idea how you found that...

f(x) = x^1.58 + 4x -320
f-1(x) = x = y^1.58 + 4y - 320...?

where do you go from there?

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transientb.bandcamp.com

The differences in terminology is killing me. What the hell is f-1(x) ?

I guess I was assuming that the "a" in your equation a^1.58... was obtained using a substitution. I didn't try the substitution myself, as I was assuming that you'd done it correctly. Then once you found "a", you could then find x.

Is the 1.58 an estimate for log(3)/log(2) ?

Ok, let me try to invoke Newton's method.

I'll take a (not so) educated guess at a = 320/4 = 80

a1 = 80 - (80^1.58 + 4(80) - 320) / (1.58*80^0.58 + 4 ) ~ 37

a2 = 37 - (37^1.58 + 4(37) - 320) / (1.58*37^0.58 + 4 ) ~ 30

a3 = 30 - (30^1.58 + 4(30) - 320) / (1.58*30^0.58 + 4 ) ~ 29

a4 = 29 - (29^1.58 + 4(29) - 320) / (1.58*29^0.58 + 4 ) ~ 29

Keep going if you want more precision, but an integer value of a is 29.

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yeah, 2^1.585x is basically 3^x

the inverse function... is f^-1(x)

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transientb.bandcamp.com

f-1 is written on paper as f^-1, but it doesn't mean f to the negetive first power, it means the inverse function. My calc teacher always says it's the worst math notation ever invented, and confuses more students than any other.

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All we learned was, whenever we take the limit and you get an answer of 0/0 or infinity/infinity, you can decide the exact limit or determine the actual indeterminate limit (zero or infinity) by taking the derivative of the top and bottom until a determinate answer appears. REALLY quick example:

x^2/x ( )

f(0) = 0/0, so there's normally NO POSSIBLE WAY (:heh) to determine the limit, right? Well there WOULD be, ahem, but it would either require the quotient rule or some nasty simplifying. So, if we use Newtons Method, we take the derivative of the top and bottom to get 2x/1. Putting zero into here would make for 0/1, or zero. Also it sometimes works and sometimes doesn't work at all, for reasons never divulged. My first example off the top of my head was
x^2 / x cos x
which is the kind of problem that Newton's method makes a lot easier, but that specific one apparently doesn't work with newton's rule

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Didnt you hear? hes actually a rubix cube now

RIP DDRKirby

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You know that actually isn't computer animated. DDRkirby has reached that level of rubixism and just offered his services for Sony.

Yeah, I remember that one, but I don't ever recall it being referred to as Newton's method even though he may have discovered it.

The rule of thumb is that it works when you need it, and (probably) doesn't when you don't. So for your example of x^2/xcosx, it cant'be used as x -> infinity since the denominator will oscillate between +/- infinity. You might be able to use it for x -> 0, but I can't really remember how to differentiate xcosx - is that where you use (udv/dx + vdu/dx)? If so, you end up with 0/1 = 0 which would make sense since it reduces to x/cosx and obviously = 0 when x = 0.

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That's not Newton's Method, that's L'Hôpital's Rule...

Oh, well Devil just corrected me good. I haven't covered that part of calc since sophomore year in high school, and I'm only just starting Calc II in college now

And I was wrong on two counts, L'Hopital's rule does indeed work on x^2 / x cosx. You would get:

2x / (cos x - xsinx)

or 0/1 towards zero, and +/- infinity (undefined) towards infinity.

But I still don't know how to solve the original problem

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Yeah I'm pretty sure you're supposed to use Newton's to solve the original problem.