Solving an Exponential Equation with Base eSolve Approximate the solution to three decimal places.
Property 3 (natural logs)
Power rule
In InDivide by 0.003.Use a calculator.The solution set is 5 1229.626 6 .Check that e0.003^1 1229.626^2 L40. NOW TRY
xL1229.626
x=
ln 40
0.003
0.003x= ln 40 e= e^1 = 1
0.003x ln e= ln 40
ln e0.003x= ln 40
e0.003x=40.
EXAMPLE 2
614 CHAPTER 10 Inverse, Exponential, and Logarithmic Functions
General Method for Solving an Exponential Equation
Take logarithms to the same base on both sides and then use the power rule
of logarithms or the special property (See Examples 1 and 2.)
As a special case, if both sides can be written as exponentials with the same
base, do so, and set the exponents equal. (See Section 10.2.)logb bx= x.
OBJECTIVE 2 Solve equations involving logarithms.We use the definition of
logarithm and the properties of logarithms to change equations to exponential form.
Solving a Logarithmic EquationSolve Give the exact solution.
Convert to exponential form.Take the cube root on each side.
Add.CHECK Original equation
LetWork inside the parentheses.Write in exponential form.✓ True
A true statement results, so the solution set is. NOW TRY
CAUTION Recall that the domain of is For this reason,
always check that each proposed solution of an equation with logarithms yields
only logarithms of positive numbers in the original equation.
y= logb x 1 0, q 2.
E- 5 + 2232 F
16 = 16
24 16
A 2232 B
3
= 23 A 232 B
3
log 2 16 4 = 8 # 2 = 16log 2 A 2232 B
(^3)
4
log 2 1 - 5 + 2232 + 523 4 x=- 5 + 2232.
log 2 1 x+ 523 = 4
x =- 5 + 2232 2316 = 238 # 2 = 238 # 232 = 2232
x=- 5 + 2316 - 5
x+ 5 = 2316
1 x+ 523 = 16 24 = 16
1 x+ 523 = 24
log 2 1 x+ 523 = 4
log 2 1 x+ 523 =4.
EXAMPLE 3
NOW TRY
EXERCISE 2
Solve. Approximate
the solution to three decimal
places.
e0.12x= 10NOW TRY
EXERCISE 3
Solve
Give the exact solution.
log 5 1 x- 123 =2.NOW TRY ANSWERS
- 5 19.188 6 3. E 1 + (^2325) F