SME
Grade 11 Math · Problem type
Equations with two logs
Use this for “Solve log₃ x + log₃(x − 8) = 2.”
Chart
Vertical asymptote x = 8
x = −1 is rejected: f(−1) has no value, it is outside the domain
f(x) = g(x)
The curves cross at x = 9.
This is an example. Type your own values to start a new problem.
log
(x)
+
log
(x
+
(
c
)
)
=
y
x(x
+
(
c
)
)
=
b
=
u
x₁
Larger candidate, the solution
worked out (not typed)
x₂
Smaller candidate
worked out (not typed)
Formulas
u
=
b
y
9
=
3
2
x₁
=
−
c
+
√
c
2
+ 4 ×
u
2
9
=
(−(−8)
+
√(
(−8)
2
+
4
×
9))
÷
2
x₂
=
−
c
−
√
c
2
+ 4 ×
u
2
−1
=
(−(−8)
−
√(
(−8)
2
+
4
×
9))
÷
2
Assumptions
•
The product rule joins the logs: logb(x) + logb(x + c) = logb(x(x + c)).
•
Then x(x + c) = bʸ, a quadratic: x² + cx − bʸ = 0.
•
A log needs a positive number, so check each candidate in x and x + c.
Step-by-step
WE KNOW
b = 3, c = −8, y = 2
FIND
the product x(x + c), bʸ (u), larger candidate, the solution (x₁), smaller candidate (x₂)
Step 1 · Find the product x(x + c), bʸ (u)
Use u = bʸ
Join
the
logs:
logb(x(x
+
c
))
=
y
,
so
x(x
+
c
)
=
bʸ.
u
=
3
2
u
=
9
Step 2 · Find larger candidate, the solution (x₁)
Use x₁ = (−c + √(c² + 4 × u)) ÷ 2
Multiply
out
x(x
+
c
)
=
u
and
use
the
quadratic
formula
on
x
2
+
cx
−
u
=
0.
x₁
=
(−(−8)
+
√(
(−8)
2
+
4
×
9))
÷
2
x
2
−
8x
−
9
=
0
x₁
=
8 +
√
100
2
x₁
=
8 + 10
2
x
=
9:
x
and
x
+
c
are
9
and
1,
both
positive
✓
x₁
=
9
Step 3 · Find smaller candidate (x₂)
Use x₂ = (−c − √(c² + 4 × u)) ÷ 2
The
same
formula
with
the
minus
sign.
Its
x
and
x
+
c
are
both
negative,
so
it
is
extraneous.
x₂
=
(−(−8)
−
√(
(−8)
2
+
4
×
9))
÷
2
x₂
=
8 −
√
100
2
x₂
=
8 − 10
2
x = −1 is not positive, so its log has no value: reject x = −1
x₂
=
−1
Check
Put the numbers back in. Both sides should match.
9
=
3
2
✓
9
=
(−(−8)
+
√(
(−8)
2
+
4
×
9))
÷
2
✓
−1
=
(−(−8)
−
√(
(−8)
2
+
4
×
9))
÷
2
✓
Topic check
See what you know: 5 questions on this topic
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