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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