Question 68752
The amount A in an account after t years of an initial principle P invested at an annual rate r compounded continuously is given by A=Pe^(rt) where r is expressed as a decimal. What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? 
A(10)=500e^(0.05(10)
A(10)=500e^0.5
A(10)=$824.36
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The amount of a radioactive tracer remaining after t days is given by 
A(t)=A(o)e^(-0.18t), where Ao is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days? 
40=Aoe^(-0.18(3))
40=Aoe^(-0.54)
Ao=40/(e^(-0.54))
Ao=40[e^(0.54))
Ao=40*1.716006862
Ao=68.64 grams at the beginning
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Find the number log5(1/5).
log(base 5) 5^(-1) = -1
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Solve loga(8x+5)=loga(4x+29)
If the logs are equal the anti-logs are equal:
8x+5 = 4x+29
4x=24
x=6
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The amount A in an account after t years from an initial principle P invested at an annual rate r compounded continuously is given by A =Pe^(rt) where r is expressed as a decimal. Solve this formula for t in terms of A, P, and r. 
For t:
e^(rt)=A/P
Take the natural log to get:
rt= ln(A/P)
t= [ln(A/P)]/r
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For P: P=A/[e^(rt)]
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For r: r=[ln(A/P)]/t
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The decibel level of sound is given by D=10 log(I/10to the -12 power), where I is the sound intensity measured in watts per square meter. Find the decibel level of a whisper at an intensity of 5.4 x 10 to the power of -10 watts per square meter. 
D=10log[(5.4 x 10^(-10))/(10^-12)
D=10log(5.4 x 10^2)
D=10log(500.4)
D=10*2.69931
D=26.99 db's
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Cheers,
Stan H.