MATH HELP!!!1

[krez]

The equation 2 = 1.0004^(n365), I need to solve for 'n'

I solved it with calculus np using natural logs, n=4.65

my question is can it be solved with algebra?  I havent done it in years and have forgotten the rules when dealing with exponents

[Barton]

Yes.

1.0004^365n = 2
-Convert to log form.
 - x^n = b -> logx(b) = n
 
log1.0004(2) = 365n
-Use the Change of Base rule to evaluate the logarithm.

log 2 / log 1.0004 = 1733.2145

1733.2145 = 365n

1733.2145/365 = n

n = 4.7485

[krez]

I meant solving it purely algebraically, with no logarithms at all.

FYI I did

ln 2 / (365 x ln 1.0004) = 4.65

[Barton]

Logarithms are algebraic, but what I see what you're saying.

[krez]

What about this one,

1000 = 900 (1+(.05/m))^(m3)

If there wasnt an 'm' in the parentheses id use just a logarithm to find the exponent

[Barton]

Sorry, I can't help you out with that problem. The problem is beyond me, if there's a way to solve it. The variable cancels itself out, when I try to solve it.

Does your textbook give an answer to this?

[haunted]

I would show the how to do this but it's always a pleasure watching a math meltdown

[krez]

no, it doesnt give an answer for that one, but in this problem, m = 6

1045 = 900 (1+(.05 / m))^(m3)

[Slayer :D]

You cannot solve directly for "m" algebraically. You will always have m in the exponent or logarithm. You can approximate m by rearranging the equation:

0 = (.05/m)^(3m) - (1/9)

Then treat it like the equation of a line:

y = (.05/x)^(3x) - (1/9)

Now we know that m = x when y = 0, so we can evaluate points above and below the x-axis. Take two very close together and calculate the equation of the line that travels through both points. Then solve for the intersection point of that line and the x-axis. The x-value of that point will closely approximate m.

[QwazyWabbit]

That should be 10/9, not 1/9.

This is not a linear equation, it's a quadratic or higher.
Did anyone notice that .05/m = 1/20 * 1/m = 1/20*m?

Not to do anyone's homework for them, but here's what the function looks like:
http://www.wolframalpha.com/input/?i=1000+%3D+900+%281%2B%28.05%2Fx%29%29%5E%283*x%29

[Barton]

That should be 10/9, not 1/9.

This is not a linear equation, it's a quadratic or higher.
Did anyone notice that .05/m = 1/20 * 1/m = 1/20*m?

Not to do anyone's homework for them, but here's what the function looks like:
http://www.wolframalpha.com/input/?i=1000+%3D+900+%281%2B%28.05%2Fx%29%29%5E%283*x%29

I don't get it. How'd you get rid of the variable exponent?

Also the answer to that is pretty off for an approximation.

[QwazyWabbit]

I was isolating the terms of Slayer's analysis. He reduced 1000/900 to 1/9, I was pointing out that it is 10/9, not 1/9.

The reduction would be 0 = (1 + 1/20*m)^3*m - 10/9 and if m is anything other than 1/3 (making 3m=1) or m=0, (making the exponent 3m=0) then the equation is not a line. He attempted to evaluate it as y = m x + b but the exponent makes it a quadratic, not a line, in this case it looks like a hyperbolic with asymptotes at 0 and ~1045 (~10.45 scaled).

[Barton]

I was isolating the terms of Slayer's analysis. He reduced 1000/900 to 1/9, I was pointing out that it is 10/9, not 1/9.

The reduction would be 0 = (1 + 1/20*m)^3*m - 10/9 and if m is anything other than 1/3 (making 3m=1) or m=0, (making the exponent 3m=0) then the equation is not a line. He attempted to evaluate it as y = m x + b but the exponent makes it a quadratic, not a line, in this case it looks like a hyperbolic with asymptotes at 0 and ~1045 (~10.45 scaled).

When I was talking about the answer, I was refer to the one given in the link you gave. What you've just said now is satisfying though.

[QwazyWabbit]

Oops. The link didn't translate correctly, amended link below:

http://www.wolframalpha.com/input/?i=1000+%3D+900+%281%2B%28.05%2Fx%29%29%5E%283x%29