Logarithmic word problem

Mar 2017
358
3
Massachusetts
Hi,

I hope someone can help me with the following question:

Screen Shot 2017-06-29 at 11.33.10 AM.png

Can someone please let me know whether my solution is correct? I said that in about in year 2015, the amount pests would be the same. The reason why I want to know whether I'm right is because my solution does not match up to the textbook's.

IMG_20170629_113620-2.jpg

I would really appreciate help!
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I get 15.93 years from 1997 ... the initial population of the Washington forest is 1.95 million, not 1.9 million
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I get a slightly different answer. I used the fact that \(\displaystyle \frac{0.97^x}{0.96^x}= \left(\frac{0.97}{0.96}\right)^x= 1.0104^x\) so that \(\displaystyle 1.0104^x= 1.2105\). Now take the logarithm of both sides: \(\displaystyle x log(1.0104)= log(1.2105)\) so \(\displaystyle 0.00500x= 0.08296\) and the \(\displaystyle x= \frac{0.08296}{0.00500}= 16.6, so the answer is 1996+ 17= 2013 rather than 2015. What was the answer in your text? It might be just a matter of round off error.\)
 
Mar 2017
358
3
Massachusetts
I get a slightly different answer. I used the fact that \(\displaystyle \frac{0.97^x}{0.96^x}= \left(\frac{0.97}{0.96}\right)^x= 1.0104^x\) so that \(\displaystyle 1.0104^x= 1.2105\). Now take the logarithm of both sides: \(\displaystyle x log(1.0104)= log(1.2105)\) so \(\displaystyle 0.00500x= 0.08296\) and the \(\displaystyle x= \frac{0.08296}{0.00500}= 16.6, so the answer is 1996+ 17= 2013 rather than 2015. What was the answer in your text? It might be just a matter of round off error.\)
\(\displaystyle

Yeah you're right, it was just a round off error. Thanks for your help!\)
 
Feb 2015
2,255
510
Ottawa Ontario
230 * (1 - .04)^n = 195 * (1 - .03)^n

Solve for n (you should get 15.93005335275....)

Remember that a^n / b^n = (a/b)^n
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
$2.3(0.96)^x = 1.95(0.97)^x$

$\dfrac{2.3}{1.95} = \dfrac{0.97^x}{0.96^x} = \left(\dfrac{0.97}{0.96}\right)^x$

$\log\left(\dfrac{2.3}{1.95}\right) = x\log\left(\dfrac{0.97}{0.96}\right)$

$\dfrac{\log \left( \frac{2.3}{1.95} \right)}{\log \left( \frac{0.97}{0.96} \right)} = x$

$15.93 = x$