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Putting these two steps together, we simplifyln(e-2t)= ln(19)to-2t=-ln(9). We arrive at our solution,t=ln(9)2which simplifies tot= ln(3).(Can you explain why?)The calculator confirms the graphs off(x) = 75 and≈1.099.y=f(x) = 9·3xandy=f(x) = 75 andy=g(x) = 72xy=g(x) =1001+3e-2x5. We start solving 25x= 5x+ 6 by rewriting 25 = 52so that we have(52)x= 5x+ 6, or52x= 5x+ 6. Even though we have a common base, having two terms on the right hand sideof the equation foils our plan of equating exponents or taking logs. If we stare at this longenough, we notice that we have three terms with the exponent on one term exactly twice thatof another. To our surprise and delight, we have a ‘quadratic in disguise’. Lettingu= 5x,we haveu2= (5x)2= 52xso the equation 52x= 5x+ 6 becomesu2=u+ 6. Solving this asu2-u-6 = 0 givesu=-2 oru= 3. Sinceu= 5x, we have 5x=-2 or 5x= 3. Since5x=-2 has no real solution, (Why not?) we focus on 5x= 3. Since it isn’t convenient toexpress 3 as a power of 5, we take natural logs and get ln (5x) = ln(3) so thatxln(5) = ln(3)
7.3 Exponential Equations and Inequalities365orx=ln(3)ln(5). When we graphf(x) = 25xandg(x) = 5x+ 6, we see that they intersect atx=ln(3)ln(5)≈0.6826.6. At first, it’s unclear how to proceed withex-e-x2= 5, besides clearing the denominator toobtainex-e-x= 10. Of course, if we rewritee-x=1ex, we see we have another denominatorlurking in the problem:ex-1ex= 10. Clearing this denominator gives use2x-1 = 10ex,and once again, we have an equation with three terms where the exponent on one term isexactly twice that of another - a ‘quadratic in disguise.’ If we letu=ex, thenu2=e2xso theequatione2x-1 = 10excan be viewed asu2-1 = 10u. Solvingu2-10u-1 = 0, we obtainby the quadratic formulau= 5±√26. From this, we haveex= 5±√26. Since 5-√26<0,we get no real solution toex= 5-√26, but forex= 5 +√26, we take natural logs to obtainx= ln(5 +√26). If we graphf(x) =ex-e-x2andg(x) = 5, we see that the graphs intersectatx= ln(5 +√26)≈2.312y=f(x) = 25xandy=f(x) =ex-e-x2andy=g(x) = 5x+ 6y=g(x) = 5The authors would be remiss not to mention that Example7.3.1still holds great educationalvalue. Much can be learned about logarithms and exponentials by verifying the solutions obtainedin Example7.3.1analytically.For example, to verify our solution to 2000 = 1000·3-0.1t, wesubstitutet=-10 ln(2)ln(3)and obtain2000?=1000·3-0.1“-10 ln(2)ln(3)”2000?=1000·3ln(2)ln(3)2000?=1000·3log3(2)Change of Base2000?=1000·2Inverse Property2000X=2000The other solutions can be verified by using a combination of log and inverse properties. Somefall out quite quickly, while others are more involved. We leave them to the reader.
366Exponential and Logarithmic FunctionsSince exponential functions are continuous on their domains, the Intermediate Value Theorem4.1applies. As with the algebraic functions in Section??, this allows us to solve inequalities usingsign diagrams as demonstrated below.