SOLUTION: divide {{{ x^5+x^3+2x+7 by x-3 }}} Solve {{{ 4^x = 12 }}} write as a single log {{{ 3 Log x - (1/3) Log y + 4 Log z }}} A radioactive isotope has a halflife of 600 years.

Algebra ->  Equations -> SOLUTION: divide {{{ x^5+x^3+2x+7 by x-3 }}} Solve {{{ 4^x = 12 }}} write as a single log {{{ 3 Log x - (1/3) Log y + 4 Log z }}} A radioactive isotope has a halflife of 600 years.      Log On


   

Question 174209: divide +x%5E5%2Bx%5E3%2B2x%2B7+by+x-3+
Solve +4%5Ex+=+12+
write as a single log +3+Log+x+-+%281%2F3%29+Log+y+%2B+4+Log+z+
A radioactive isotope has a halflife of 600 years. If 50g is on hand now, how much will be available in 1000 years?


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
divide +x%5E5%2Bx%5E3%2B2x%2B7+by+x-3+
Use synthetic division:
3)....1....0....1....0....2....7
.......1....3....10...30...92..|..283
Quotient: x^4 + 3x^3 + 10x^2 + 30x + 92
Remainder: 283
==========================================
Solve +4%5Ex+=+12+
xlog(4) = log(12)
x = log(12)/log(4) = 1.7925
===========================================
write as a single log +3+Log+x+-+%281%2F3%29+Log+y+%2B+4+Log+z+
log(x^3) - log(y^(1/3)) + log(z^4)
= log[x^3*z^4/y^(1/3)]
============================================
A radioactive isotope has a halflife of 600 years. If 50g is on hand now, how much will be available in 1000 years?
---
A(t) = Ao (1/2)^(t/600)
A(1000) = 50*(1/2)^(1000/600)
A(1000) = 50*0.31498
A(1000) = 15.749 grams
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Cheers,
Stan H.