Exponential and Log Equations

Equations that Involve Logs
Step by Step Method
Step 1: Contract to a single log.
Step 2: Get the log by itself.
Step 3: Exponentiate both sides with the appropriate base.
Step 4: Solve.
Step 5: Check your solution for domain errors.
Example:
Solve
log_{5} x + log_{5} (x + 2) = log_{5} (x + 6)

log_{5} x + log_{5} (x + 2)  log_{5}
(x + 6) = 0
log_{5} x (x + 2)  log_{5} (x + 6) = 0
log_{5} x (x + 2)/(x + 6) = 0

Already done.

x(x + 2)/(x + 6) = 5^{0} = 1

x(x + 2) = x + 6
x^{2} + 2x  x  6 = 0
x^{2} + x  6 = 0
(x  2)(x + 3) = 0
x = 2 or x = 3

Note that 3 is not in the domain of the first log hence the only
solution is x = 2.
Exercises: Solve

log(x + 6) + 1 = 2log(3x  2)

1/2 log(x + 3) + log2 = 1

Exponential Equations
Step 1: Isolate the exponential
Step 2: Take the appropriate log of both sides.
Step 3: Solve
Example: Solve
4e^{7x} = 15

e^{7x} = 15/4

lne^{7x} = ln(15/4)

7x = ln(15/4)

x = ln(15/4)/7
Exercises:
Solve

1 + 2e^{x} = 9

(10^{x  4})/e^{2x}  4 = 0

(lnx)^{2} = ln(x^{2})

2^{3x} + 4(2^{3x}) = 5

Application
All living beings have a certain amount of radioactive carbon C_{14}
in their bodies. When the being dies the C_{14} slowly decays
with a half life of about 5600 years. Suppose a skeleton is found in
Tahoe that has 42% of the original C_{14}. When did the
person die?
Solution:
We can use the exponential decay equation:
y = Ce^{kt}
After 5600 years there is
C/2
C_{14} left. Substituting, we
get:
C/2 = Ce^{k(5600)
}
Dividing by C,
1/2 = e^{5600k
}Take ln of both sides,
ln(.5) = 5600k
so that
k = [ln(.5)]/5600 = .000124
The equation becomes
y = Ce^{.000124t}
To find out when the person died, substitute
y = .42C
and solve for t:
.42C = Ce^{.000124t
}
Divide by C,
.42 = e^{.000124t
}
Take ln of both sides,
ln(.42) = .000124t
Divide by .000124
t = [ln(.42)]/(.000124) = 6995
The person died about 7,000 years ago.
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