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Jim Lamb ers Math 1B Fall Quarter 2004- Le ture 6 Notes
These notes orresp ond to Se tion 4.5 in the text.
Exp onential and Logarithmi Equations
Now that we have dis ussed exp onential fun tions and well as their inverses, logarithmi fun tions, we are able to solve equations that involve either logarithms or exp onents. In this le ture, we dis uss some te hniques for solving su h equations.
Exp onential Equations
To solve an exponential equation for an unknown x, where x o urs within exp onents, a useful approa h is to rearrange the equation so that it has the form
y = af^ (x)^ ; (1)
where f (x) is some fun tion of x. Then, we an take the logarithm, to some base b, of b oth sides and use the prop erties of logarithms to obtain
f (x) = log (^) b y log (^) b a
Then, we an try to solve this equation, whi h is likely to b e mu h simpler than the original equation. Any p ositive numb er b (other than one) an b e used for the base; typi ally the hoi e is di tated by onvenien e, su h as setting b = a so that log (^) b a = log (^) a a = 1.
Example 1 Supp ose that a sample of a radioa tive substan e de ays from 100 g to 10 g in ve days. We will ompute the half-life of the substan e. Re all that radioa tive de ay is mo deled by the equation
A = A 0
t=h ; (3)
where A 0 is the initial amount of the substan e, A is the amount after time t has elapsed, and h is the half-life. Substituting A 0 = 100, A = 10, and t = 5, we have
10 = 1002 ^5 =h^ ; (4)
where we have used the fa t that (1=2)x^ = 1 x^ = 2 x^ = 2 x^. Rearranging yields the equation
10 ^1 = 2 ^5 =h^ : (5)
Sin e the base on the left side is 10, we take the ommon logarithm of b oth sides and use the prop erties of logarithms to obtain
log 10 =
h log 2 : (6)
Rearranging, we obtain h = 5 log 2 1 : 505 ; (7)
sin e log 10 = 1. We on lude that the half-life of the substan e is approximately 1.505 days. 2
Another approa h is useful in ases where the equation an b e written in su h a way that it ontains expression of the form bf^ (x)^ , b^2 f^ (x)^ , b^3 f^ (x)^ , et ., with no other o urren es of x in the equation. Then, we an make a substitution u = bf^ (x)^ , whi h yields b^2 f^ (x)^ = u^2 , b^3 f^ (x)^ = u^3 , and so on, in view of the prop erty axy^ = (ax^ )y^. The result is a p olynomial in equation in u, whi h we then solve for u. The nal step onsists of solving the equation u = f (x) for x.
Example 2 We will solve the equation
ex^ + e x^ = 2 (8)
for x. If we multiply through by ex^ , we obtain
e^2 x^ + 1 = 2 ex^ : (9)
Now, we use the substitution u = ex^ , whi h yields the quadrati equation
u^2 2 u + 1 = 0 : (10)
The left side of this equation fa tors into (u 1)^2 , whi h yields u = 1. To obtain x, we solve the equation u = ex^ for x. Sin e u = 1, we have ex^ = 1. Taking the natural logarithm of b oth sides, we obtain ln ex^ = ln 1 : (11)
By the prop erties of logarithms, this simpli es to
x ln e = 0 ; (12)
and sin e ln e = 1, we on lude that x = 0. 2
We b egin by onverting the equation log (^) b y = x, where y is known, from logarithmi form to exp onential form. This yields the exp onential equation
bx^ = y : (19)
To solve this equation, we pro eed as des rib ed earlier in this le ture, by taking the natural loga- rithm of b oth sides. Using the prop erties of logarithms, we obtain
x ln b = ln y ; (20)
whi h yields the solution
x = ln y ln b
We have just derived the hange-of-base formula
log (^) b y =
ln y ln b
Example 4 To ompute log 5 10, we apply the hange-of-base formula, whi h yields
log 5 10 = ln 10 ln 5
