# population word problems logarithmic

0(The \({{\log }_{2}}2\) has an invisible plus before it). Remember from Parent Graphs and Transformations that the critical or significant points of the parent logarithmic function \(y={{\log }_{b}}x\) (inverse of exponential function) are \(\displaystyle \left( {\frac{1}{b},\,-1} \right),\,\left( {1,0} \right),\,\left( {b,1} \right)\). (b) How many years will it take for the population to double? You have a seismograph (fourdigityear(now.getYear())); (a) What will be the population in 2015? Genius! Since the whole term is raised to 4, first move the 4 down around to the front. eval(ez_write_tag([[728,90],'shelovesmath_com-medrectangle-3','ezslot_2',109,'0','0']));A slide rule was used (among other things) to multiply and divide large numbers by adding and subtracting their exponents. We will also discuss why the base of e is used so often with population problems. A large colony of fleas is growing exponentially on the family dog (yuck!). Therefore, you must read this article “Real Life Application of Logarithms” carefully. = 0.0003, so: pH = log[H+] On to Solving Inequalities – you are ready! Note: You might be asked to solve simple problems like these using the log “loop” and change of base formula instead: Solving Logs: Use the Log “Loop” and Change of Base to get Exponent Down, \(\displaystyle \begin{align}{{\left( {.5} \right)}^{t}}&=\frac{{1500}}{{150}}=10\\{{\log }_{{.5}}}\left( {10} \right)&=t\\t&={{\log }_{{.5}}}\left( {10} \right)\\t&=\frac{{\log \left( {10} \right)}}{{\log \left( {.5} \right)}}\approx -3.32\end{align}\), \(\begin{align}{{\log }_{5}}\left( {1000} \right)&=x-2\\x&={{\log }_{5}}\left( {1000} \right)+2\\x&=\frac{{\log \left( {1000} \right)}}{{\log \left( 5 \right)}}+2\approx 6.29\end{align}\). If \(x\) is underneath a complicated exponent, raise each side to the reciprocal of that exponent. When functions are transformed on the outside of the \(f(x)\) part, you move the function up and down and do “regular” math, as we’ll see in the examples below. (Note: for \(y={{\log }_{3}}\left( {2\left( {x-1} \right)} \right)-1\), for example, the \(x\) values for the parent function would be \(\displaystyle \frac{1}{3},\,\,1,\,\,\text{and}\,\,3\). accessdate = date + " " + so this is basic. page, Logarithmic Now we’re going to use these properties to expand and condense logs. event, I need to convert the intensity to a Richter rating by evaluating (See the plot in Figure 1). Both work! | 1 | 2 | 3 For example, if we start out with 20 grams, after the next time period, we’d have 10, then 5, and so on. (a) We’ll use the exponential decay formula, where \(P\), (a) To get the population in year 2015, we can get \(t\) by subtracting 2009 from 2015: to get \(t=6\). \(\displaystyle {{32}^{x}}=\frac{1}{4},\,\,\,\,\,{{\left( {{{2}^{5}}} \right)}^{x}}={{2}^{{-2}}},\,\,\,\,\,\,{{2}^{{5x}}}={{2}^{{-2}}},\,\,\,\,5x=-2,\,\,x=-\frac{2}{5}\). \(\displaystyle \begin{align}{{\log }_{2}}{{4}^{4}}+{{\log }_{2}}{{x}^{{\frac{1}{2}}}}&=\,{{\log }_{2}}\left( {{{4}^{4}}{{x}^{{\frac{1}{2}}}}} \right)\,\,\,\\\,\,\,\,\,\,\,&=\,{{\log }_{2}}\left( {256\sqrt{x}} \right)\end{align}\), Apply the power rule first by moving up the coefficients and making them exponents. endobj Logarithmic word problems, in my experience, generally involve evaluating a given logarithmic equation at a given point, and solving for a given variable; they're pretty straightforward. We’ll soon see that Logs can be used to “get the variable in the exponent down” so we can solve for it. 2. Given states the information given by the problem. \(\displaystyle \begin{align}p\left( t \right)&=50{{e}^{{.085t}}}\\p\left( 6 \right)&=50{{e}^{{.085\left( 6 \right)}}}\\&\approx 83.26\\&=83\text{ wolves}\end{align}\). = 1.3 × 109. Then, simplify the \({{\log }_{2}}2\) to 1. It’s a good idea to store \(k\) in our calculator again. From counting through calculus, making math make sense! The beginning amount \(P\) is still 40, the ending amount \(A\) is 2, and \(k\) is the same. There are 400 fleas initially. The population of a town is declining by 3% every year. So when you take the log of something, you are getting back an exponent. You typically use this if you have variables raised to exponents. (Population Word Problems) To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation. eval(ez_write_tag([[336,280],'shelovesmath_com-leader-3','ezslot_12',112,'0','0']));Again, we typically use logs to solve problems where we have a variable in the exponent; we can use the Power Rule to “get the exponent down”. \(\displaystyle \begin{align}y&=a{{\left( {.5} \right)}^{{\frac{t}{p}}}}\\2&=\left( {40} \right){{\left( {.5} \right)}^{{\frac{t}{6}}}}\\.05&={{.5}^{{\frac{t}{6}}}}\\\ln \left( {.05} \right)&=\ln {{\left( {.5} \right)}^{{\frac{t}{6}}}}\\\ln \left( {.05} \right)&=\frac{t}{6}\ln \left( {.5} \right)\\t&=\frac{{\ln \left( {.05} \right)}}{{\ln \left( {.5} \right)}}\left( 6 \right)\\&\approx 25.93\end{align}\), \(\require{cancel} \displaystyle \begin{align}A&=P{{e}^{{kt}}}\\1&=2{{e}^{{k(6)}}}\\.5&={{e}^{{k(6)}}}\\\ln \left( {.5} \right)&=\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{k(6)}}}} \right)\\6k&=\ln \left( {.5} \right)\\k&=\frac{{\ln \left( {.5} \right)}}{6}\\&\approx -.1155245301\\\\A&=40{{e}^{{-.1155245301t}}}\\&=40{{e}^{{-.1155245301\left( {18} \right)}}}\\&=5\text{ grams}\end{align}\), (a) We want to solve for \(k\) first in the exponential equation \(A=P{{e}^{{kt}}}\) (which is also seen as \(\displaystyle N\left( t \right)={{N}_{0}}{{e}^{{kt}}}\)) when, \(\displaystyle \begin{align}A&=P{{e}^{{kt}}}\\2&=40{{e}^{{-.1155245301t}}}\\.05&={{e}^{{-.1155245301t}}}\\\ln \left( {.05} \right)&=\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{-.1155245301t}}}} \right)\\t&=\frac{{\ln \left( {.05} \right)}}{{-.1155245301}}\\&\approx 25.93\end{align}\), (b) Now we need to use logs again to find the time until there are only, Note that we have to solve the log inequality, but also set the. I have to admit that logs are one of my favorite topics in math. Remember to always check your answer to make sure the argument of logs (what’s directly following the log) is positive! Let’s do some problems and see what techniques we use: \(\begin{align}2{{n}^{2}}+10&=22-5n\\2{{n}^{2}}+5n-12&=0\\\left( {2n-3} \right)\left( {n+4} \right)&=0\\n=\frac{3}{2},\,\,\,\,n=-4\end{align}\). 2 0 obj Everything with a \(+\) sign before it goes on top (numerator), and with a – sign before it goes on bottom (denominator). Since everything is under the root, first move the around to the front (the 4th root means raised to the \(\displaystyle \frac{1}{4}\)). When working with logs, there are certain shortcuts that you can use over and over again. var now = new Date(); should you follow the rules and wear ear protection when relaxing at the Richter function at I 1 0 obj Note: If all else fails, you can solve log equations using a graphing calculator; here’s an example: Push GRAPH. decibels can cause hearing damage or loss, and considering that a gunshot Since this is well above the level at which I can suffer hearing Population Growth Problem. We will get the same answer.). Remember again that everything on top has a plus before it, and everything on bottom has a minus before it.

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