This is the given equation add 3 to both sides

This is the given equation.
Add 3 to both sides.
Isolate the exponential factor by dividing both
sides by 40.
Take the natural logarithm on both sides.
Use the inverse property on the left.
Divide both sides by 0.6 and solve for x .
x =
ln 6
0.6
L 2.99
ln e
x
= x
0.6 x = ln 6
ln e
0.6 x
= ln 6
e
0.6 x
= 6
40 e
0.6 x
= 240
40 e
0.6 x
- 3 = 237
e
0.6 x
.
40 e
0.6 x
- 3 = 237.
EXAMPLE 3
Thus, the solution of the equation is Try checking this approximate
solution in the original equation to verify that is the solution set.
Check Point
3
Solve: Find the solution set and then use a
calculator to obtain a decimal approximation to two decimal places for the solution.
Solving an Exponential Equation
Solve:
Solution
Because each exponential expression is isolated on one side of the
equation, we begin by taking the natural logarithm on both sides.
This is the given equation.
Take the natural logarithm on both sides.
Use the power rule and bring the variable
exponents to the front:
Use the distributive property to distribute
ln 5 and ln 4 to both terms in parentheses.
Collect variable terms involving on the
left by subtracting and adding 2 ln 5
on both sides.
Factor out from the two terms on the left.
Isolate by dividing both sides by
The solution set is The solution is approximately
Check Point
4
Solve: Find the solution set and then use a
calculator to obtain a decimal approximation to two decimal places for the solution.
3
2 x - 1
= 7
x + 1
.
- 6.34. e
2 ln 5 + 3 ln 4
ln 5 - 2 ln 4
f .
ln 5 - 2 ln 4.
x
x =
2 ln 5 + 3 ln 4
ln 5 - 2 ln 4
x
x 1 ln 5 - 2 ln 4 2 = 2 ln 5 + 3 ln 4
2 x ln 4
x
x ln 5 - 2 x ln 4 = 2 ln 5 + 3 ln 4
x ln 5 - 2 ln 5 = 2 x ln 4 + 3 ln 4
ln b
x
= x ln b .
(x-2) ln 5=(2x+3) ln 4
Be sure to insert parentheses around the binomials.
Remember that ln 5 and ln 4 are constants, not variables.
ln 5
x - 2
= ln 4
2 x + 3
5
x - 2
= 4
2 x + 3
5
x - 2
= 4
2 x + 3
.
EXAMPLE 4
7 e
2 x
- 5 = 58.
e
ln 6
0.6
f
ln 6
0.6
L 2.99.
Discovery
Use properties of logarithms to show
that the solution in Example 4 can be
expressed as
ln 1600
ln
A 5
16
B
.

Section 3.4 Exponential and Logarithmic Equations
427
Technology
Graphic Connections
Shown below is the graph of
There are two
one at 0 and one at
approximately 1.10. These intercepts
verify our algebraic solution.
[ − 3, 3, 1] by [ − 1, 3, 1]
x -intercepts,
y = e
2 x
- 4 e
x
+ 3.
Solving an Exponential Equation
Solve:
Solution
The given equation is quadratic in form. If the equation can be
expressed as Because this equation can be solved by factoring, we
factor to isolate the exponential term.
This is the given equation.
Factor on the left. Notice that if
Set each factor equal to 0.
Solve for
Take the natural logarithm on both sides
of the first equation. The equation on
the right can be solved by inspection.
The solution set is The solutions are 0 and ln 3, which is approximately
1.10.
Check Point
5
Solve: Find the solution set and then use a
calculator to obtain a decimal approximation to two decimal places, if necessary,
for the solutions.