document.write( "Question 427022: I have been reading and trying very hard to get these logarithms, they just make my head spin. I am starting to kind of get the basics, but the book my school uses does not explain things clearly. How would one go about graphing these? Please show all steps so I can try to understand this process from start to finish.\r
\n" );
document.write( "\n" );
document.write( "A) y=1/2e^x-1\r
\n" );
document.write( "\n" );
document.write( "B) y= -3 log x + 2 \n" );
document.write( "
Algebra.Com's Answer #297356 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! I have been reading and trying very hard to get these logarithms, they just make my head spin. I am starting to kind of get the basics, but the book my school uses does not explain things clearly. How would one go about graphing these? Please show all steps so I can try to understand this process from start to finish. \n" ); document.write( "A) y=1/2e^x-1 \n" ); document.write( "B) y= -3 log x + 2 \n" ); document.write( ".. \n" ); document.write( "Let us take the first equation, y=1/2e^x-1 \n" ); document.write( "Let us break it into its parts. \n" ); document.write( "First, let y=1/e^x \n" ); document.write( "You can see that when x=0, y=1. \n" ); document.write( "This will be true no matter what the base is. \n" ); document.write( "You can also see that as x becomes very large, y approaches zero but does not reach zero. This makes the x-axis an asymptote. \n" ); document.write( "Now we have two things to graph the equation, the point (0,1) and the x-axis as the asymptote. \n" ); document.write( "What happens to the curve when x<0? \n" ); document.write( "The equation will now change from 1/e^-x to e^x. As you can see as x becomes more negative, e^x becomes larger and larger to infinity. So now you should have enough information to graph 1/e^x. But wait we want to graph 1/2e^x-1. No sweat, all we need is a little more algebra. The 2 in the denominator changes the y-intercept from (0,1) to (0,1/2) and makes the curve steeper. The -1 just bumps the entire curve down one unit. This changes the asymptote to y=-1 instead of the x-axis. \n" ); document.write( "That is all there is to it. \n" ); document.write( ".. \n" ); document.write( "Let us try the second equation, y= -3 log x + 2 \n" ); document.write( "first change -3logx to logx^-3 or log 1/x^3 \n" ); document.write( "When x=1,log 1/x^3=log 1/1=log1=0 (Log 1 of any base =0) \n" ); document.write( "So you have this point to graph, (1,0) \n" ); document.write( "As x becomes less than 1, 1/x^3 increases rapidly approaching the y-axis but does not cross the y-axis because there is no such thing as the log of a negative number. So the y-axis becomes an asymptote for the graph. As x becomes greater than 1, you are not taking the logarithim of a fraction which gives you a negative number, so y will become increasingly negative but not as rapidly. The+2 will just bump the entire curve up 2 points. \n" ); document.write( "Hope this helps! \n" ); document.write( "See the graph of the first equation below: The second equation will not graph with this format.\r \n" ); document.write( "\n" ); document.write( ".. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |