dating timelines

The Panic Years by Doree Lewak

dating fossils wiki

How did they date the fossils in the geologic column?

http://en.wikipedia.org/wiki/Charles_Lyell

The fossils had already been arranged in the geologic column by the time Charles Lyell wrote his book (Principles of Geology early 1800s).

Charles Darwin took a copy of his book with him on The Beagle.

They didn’t have our sophisticated dating tools back then.

I think they did it by size. The smallest ones were first. They were 550 million years old. The next size up were next, and so on and so on . . .

but the dinosaurs couldn’t come before humans so they had to adjust their scale for that.

Evolution and the age of Earth

carbon 14 dating calculator

need some help to make sure for some of my answers and explaination to others ..?

(>) is for my answers
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1. ln(e^1.618) =
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>1.618 is there any explaination to how to find it without using the calculator??
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2. Sharon invest $2500 at an annual interest rate 9%. how much is tl investment worth after 10 years if the inter is compounded continuously?
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>i use this formula A=Pe^rt, =614.9
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3. The age of an artifact can be determined using carbon-14 dating with the equation N(t)=N(0)e^-0.00012t. What is the approximate age of an artifact if a sample reveals that it contains 34% of its original carbon-14?
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>i didnt understand this question?!?!
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4.log(x)1/128=-7
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>x=2
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5.xlog1/6=log6
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>x=-1
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6.ln(2x)=4ln2
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>x=8
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7.lnx^0.5 - 3=1
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>?????
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8. ln(x+1)=2ln4
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>x=15
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9. logx+log(x+3)=1
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>????
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10.3^x=5^2.3
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>x=3.3695
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thank you very much for your time, i really appreciate it

1) ln.x and e^x are the opposites of each other - so ln.(e^x) = x.
Just like if I said what is x * (8 / x) you know the answer is 8 because dividing by x is the opposite of multiplying by x.

2) Your answer must be wrong, as you’ve ended up with less money than she started with ($2500)! And you’d expect that after investing for 10 years at a good rate of interest (9%) you’d have more money. So that should make you think and check your method.
The formula you’ve written is wrong - you don’t need e in the equation.
I would use this equation:
A = P * r^t
P is the original investment, 2500.
r is 9% which as a multiplication factor is 1.09 (because after each year, you add 9%, so you have 109% total compared to the start, and 109% = multiply by 1.09).
t is 10 (years).
So A = 2500 * (1.09 ^ 10) = 2500 * 2.367 = $5918

3) Well, you have the formula, you just need to work out how the information in the question relates to it. N(t) is the amount of C14 left after t years. N(0) is the amount you had t the start. You’re told that N(t) is 34% of N(0), so if we divide both sides of the equation by N(0) we get:
N(t) / N(0) = 34% = 0.34 = e^-0.00012t
Take the natural log of each side:
ln0.34 = -1.0788 = ln(e^-0.00012t) = -0.00012t
So t = -1.0788 / -0.00012 = 8990 years.

4), 5) and 6) are all correct.

7) Add 3 to both sides, and remember that lnx^y = ylnx
So:
0.5 ln x = 1+3 = 4
lnx = 8
x = e^8 = 290.96

and 10) are correct.

9) remember ln(A + B) = lnA*lnB
So logx + log(x+3) = logx + logxlog3 = logx(1 + log3) = 1
I presume you’re using the convention that log means log (base 10) - if not you’ll have to correct the answers from here on
log x = 1/(1 + log3) = 1/(1 + 0.4771) = 0.6770
So x = 10^0.6770 = 4.75

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