Hi wonderful mathophles:
From an recent NYS Algebra 2 Regents, a typical question about different rates of population growth in two towns, with different starting populations.
The question was in approximately how many years will the populations be equal?
It was easy enough to just plug in the answers and see which worked, as it was multiple choice, and Desmos gives you the answer graphically in seconds.
But can it be solved using logarithms or some other method? If so how?
Here is the system: (the first number on each side is the starting population; the base of the exponential portion is the growth rate.)
1240*1.06^x = 890*1.11^x, where x is the number of years.
I assume this is elemental to many of you, but thanks for your time.
VJ
$1240\cdot 1.06^x = 890\cdot 1.11^x$
Yes this can abolutely be solved by logarithms.
I would be inclinded to do this little bit of algebra
$\frac {1240}{890} = (\frac {1.11}{1.06})^x$
before taking the log of both sides. It it not actually necessary, though.
$\log\frac {1240}{890} = \log (\frac {1.11}{1.06})^x = x\log\frac {1.11}{1.06}\\ x = \dfrac {\log\frac {1240}{890}}{\log\frac {1.11}{1.06}}= \frac {\log 1240 -\log 890}{\log 1.11 - \log 1.06}$
You could just take the logs straight away.
$\log (1240\cdot 1.06^x) = \log (890\cdot 1.11^x)$
we still have to remember our rules of the algebra of logarithms.
$\log 1240 + \log 1.06^x = \log 890 + \log 1.11^x\\ \log 1240 + x\log 1.06 = \log 890 + x\log 1.11\\ \log 1240 - \log 890 = x(\log 1.11 - \log 1.06)\\ x = \frac {\log 1240 - \log 890}{\log 1.11 - \log 1.06}$