Solving Equations with e and ln x - mit.edu

1 Solving Equations witheandlnxWe know that the natural log function ln(x) is defined so that if ln(a) =btheneb=a. Thecommon logfunction log(x) has the property that if log(c) =dthen10d=c. It s possible to define a logarithmic function logb(x) for any positivebasebso that logb(e) =fimpliesbf=e. In practice, we rarely see bases otherthan 2, 10 fory:1. ln(y+ 1) + ln(y 1) = 2x+ lnx2. log(y+ 1) =x2+ log(y 1)3. 2 lny= ln(y+ 1) +xSolve forx(hint: putu=ex, solve first foru) +e xex e x= +e xSolutions1. ln(y+ 1) + ln(y 1) = 2x+ equation involves natural logs. We apply the inverseexof the func-tion ln(x) to both sides to undo the natural (y+ 1) + ln(y 1) = 2x+ lnxeln(y+1)+ln(y 1)=e2x+lnxeln(y+1) eln(y 1)=e2x elnx(y+ 1) (y 1) =e2x xy2 1 =xe2xy2=xe2x+ 1y= xe2x+ 1We know that we cannot take the natural log of a negative number (orof 0), and our equation contains the expression ln(y 1).

2 Therefore, thesolutiony= xe2x+ 1 is not valid. Our final solution is:y= xe2x+ log(y+ 1) =x2+ log(y 1).This equation involves the log base 10, so we apply the inverse function10xto both sides. If we wished, we could subtract log(y 1) from bothsides before doing so; the result is the (y+ 1) =x2+ log(y 1)10log(y+1)= 10x2+log(y 1)y+ 1 = 10x2 10log(y 1)y+ 1 = 10x2 (y 1)y+ 1 = 10x2y 10x2y 10x2y= 1 10x2y(1 10x2) = 1 10x2y= 1 10x21 10x2 1 1y=10x2+ 110x2 1It s a good idea to check our work by pluggingy=10x2+110x2 1back into theoriginal 2 lny= ln(y+ 1) + again, we apply the inverse functionexto both sides. We coulduse the identitye2 lny= (elny)2or we could handle the coefficient of 2 asshown lny= ln(y+ 1) +xlny2= ln(y+ 1) +xelny2=eln(y+1) exy2= (y+ 1) exy2 ex y ex= 0 This is a second degree polynomial iny; the fact that some of the coeffi-cients are functions ofxshould not slow us down.

3 Applying the quadraticformula we get:y=ex ( ex)2 4 1 ( ex)2 1y=ex e2x+ original equation is valid only fory >0, and e2x+ 4ex> e2x=ex,so our final answer is:y=ex+ e2x+ best way to check our work here might be to choose some simple valuesforxand evaluate both sides of the original equation using a +e xex e x= start by applying the hint, lettingu= +e xex e x=yex+1exex 1ex=yu+1uu 1u=yu+1uu 1u uu=yu2+ 1u2 1=yu2+ 1 =y(u2 1)u2 yu2= y 1u2(1 y) = (y+ 1)u2=y+ 1y 1u= y+ 1y 1 Becauseu=exis always positive, we now have:u=ex= y+ 1y the previous problems, the variable we were Solving for was part of theinput to a logarithmic function; we isolated it by using the exponentialinverse of that logarithmic function.

4 In this problem our variable is theinput to an exponential function and we isolate it by using the logarithmicfunction with the same y+ 1y 1ln(ex) = ln( y+ 1y 1)x= ln[(y+ 1y 1)12]3x=12ln(y+ 1y 1)x=12(ln(y+ 1) ln(y 1))There are many equivalent correct answers to this question. The bestanswer is the one that is easiest for you to use and is relatively simple to check thatex= y+1y 1is correct by plugging into the original equation. We might check our final answer by plugging itinto this equation rather than the that our solution only works fory >1. Ify < 1 we can substitutev= xto see that:ev+e vev e v= y > identical calculation then yields:v=12(ln(( y) + 1) ln(( y) 1))x= 12(ln( y+ 1) ln( y 1)).

5 +e we begin by applying the hintu=ex. We solve forueither bycompleting the square or by using the quadratic +e xy=u+1uy u=u2+ 1u2 yu+ 1 = 0u=y ( y)2 4 1 12 1u=y y2 42We now replaceubyexand use the inverse function lnxto complete y2 42ex=y y2 42x= ln(y y2 42)x= ln(y y2 4) ln(2)4