§4.5 Exponential and Logarithmic Equations
Equations involving exponential end logarithmic function are called ex-
ponential and logarithmic equations.
Example 1 Solving equations for xto four decimal places.
(A). 23x−2= 5 (B). log(x+ 3) + log x= 1
Solution. (A). First use logarithm function to get xout of the exponent,
and then solve the equation for x.
23x−2= 5
log 23x−2= log 5 logarithm function is one to one.
(3 x−2) log 2 = log 5 Use logbNp=plogbNto get 3 x−2
out of the exponent position.
3x−2 = log 5
log 2
3 = 1
3µ2 + log 5
log 2 ¶Remember : log 5
log 2 6= log 5 −log 2.
= 1.4406 To four decimal places.
(B). First use properties of logarithms to express the left side as a single
logarithm, then convert to exponential form and solve for x.
log(x+ 3) + log x= 1
log[x(x+ 3)] = 1 Combine left side using log M+ log N= log M N
x(x+ 3) = 101Change to equivalent exponential form
x2+ 3 x−10 = 0
(x+ 5) (x−2) = 0
x=−5 or x= 2
Check:x=−5: log(−5 + 3) + log(−5) is not defined because the domain
of the log function is (0,∞).
x= 2: log(2 + 3) + log 2 = log 5 + log 2 = log(5 ×2) = log 10 = 1.
Thus, the only solution to the original equation is x= 2. ¤
Why we obtain an additional solution x=−5that is not a solution of the
original equation. Because in the original equation, we need x+ 3 >0and
x > 0. However, in the equation
log[x(x+ 3)] = 1
we only need the condition that x(x+ 3) >0. Hence, the range of xis
enlarged.
Example 2: Compound InterestHow many years to the nearest year
will it take the money to double if it is invested at 6% compounded annually?
Solution. Let P0be the amount of principal money. Then the amount of
money P(t) in the account after tyears, assuming no withdrawals, is given
by
P(t) = P0(1 + r
n)nt =P0(1 + 0.06
1)1×t=P0(1.06)t.
1
