Solving Equations with e and ln x

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.

3 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. 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.

4 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. 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.

5 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)). +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