Help with rearranging formulae

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Graemermac

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Hello, looking for a bit of help with these...

ln (y + 5) = 3.1, solve for Y to three decimal places

4 (^x+1) = 6^0.5x, solve for x

I= Imax (1 -e^-30t/0.01), find t

any help would be greatly appreciated, thanks
 
Well then, maybe you should spend less time on the web and more time reading your assignments!

You missed the point that this isn't the original poster.

However, having been down the pub watching Australia-v-someone . I shall attempt to answer the original questions. Probably wrongly.

OK, first one,

ln (y + 5) = 3.1, solve for Y to three decimal places

Well, I think that is,
Take antilog of both sides (is that the right term?)

(y+5) = e^3.5
Y=e^3.5-5
Y=17+ a bit I think.

Second one,
Not a valid equation. I think?

Third one,
Don't do Maximums.

Mike.
 
Probably wrongly.

Hopefully so -- maybe that will teach someone a lesson Not me of course! As you pointed out, I am not the OP, but a random member who finds it mildly annoying that the whippersnappers these days are conscripting people on the interwebs to do their homework for them.

Michael
 
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a random member who finds it mildly annoying that the whippersnappers these days are conscripting people on the interwebs to do their homework for them.

I must say, this thread wasn't even dimly disguised as anything other than a piece of homework hopefully Hero's advice will be enough for them to learn how to do their own homework rather than cheating
 
Well, one week after the original post the homework should have already been handed in. So, who's up to answering questions 2 and 3?

Mike.
 
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No thanks, I'm glad that the days of calculus homework are behind me now also, the second question seems to have incorrect notation. I assume it's meant to be "4^(x+1) = 6^0.5x, solve for x"
 
t=0.01(ln(I/Imax-1))/30

Did it in my head, so double check me.

2nd wasn't a valid expression.
 
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4^(x+1) = 6^0.5x, solve for x

(x+1)log4=.5xlog6
x+1/.5x = log6/log4
2 + 2/x = log2
x = 2(1/log2-2)

Might be wrong, been 20years since college.
 
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