Chapter Nineteen — Exponential Functions
The Humongous Book of Algebra Problems
431
19.32 Solve the equation for x: log
2
(x + 3) – log
2
9 = 4.
Express the difference of the logarithms left of the equal sign as the logarithm
of the quotient of the arguments.
Eliminate the logarithm from the equation by exponentiating
both sides of the equation with base 2.
Note: Problems 19.33–19.34 refer to the equation (ln
x)
2
– 2
ln
x
3
= –5.
19.33 Solve the equation for x.
Apply a logarithmic property to express –2
ln
x
3
as –6
ln
x.
(ln
x)
2
– 6
ln
x = –5
Solve the equation by factoring.
The solution to the equation is x = e or x = e
5
.
If you’re
taking the log
of something to
a power, you can
pull that power out
in front of the log as
a coefcient. This
log already has a
coefcient of –2, so when
you pull the exponent 3
out front, multiply it by
the coefcient that’s
already there:
–2
ln
x
3
= –2
˙
3
˙
ln
x
= –6
ln
x.
Factoring
this works
just like factoring
a regular quadratic
polynomial: w
2
– 6w + 5 =
(w – 5)(w – 1). The only
difference is that
w = ln
x.
Chapter Nineteen — Exponential Functions
The Humongous Book of Algebra Problems
432
Note: Problems 19.33–19.34 refer to the equation (ln
x)
2
– 2
ln
x
3
= –5.
19.34 Verify the solution(s) generated in Problem 19.33.
Substitute x = e and x = e
5
into the equation, one at a time, and verify that both
of the resulting statements are true.
19.35 Solve the equation for x: log
(x
2
+ 1) = 1.
Eliminate the logarithm property to rewrite the left side of the equation as a
single logarithm.
Eliminate the logarithm by exponentiating both sides of the equation with
base 10.
Solve for x.
Both x = –3 and x = 3 are valid solutions.
ln
x and e
x
cancel each
other out, leaving
behind only the
exponent of e. When
something has no
exponent written, the
implied exponent is
1, so (ln
e)
2
=
(ln
e
1
)
2
= (1)
2
.
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