1.Algebra Booster

3.2 Algebra Booster
3. LOGARITHMIC EQUATION
Type 1: A logarithmic equation is of the form
log g(x)
f(x) = b
fi f(x) = g(x) b , g(x) > 0, g(x) π 1
Type 2: A logarithmic equation is of the form
log f1() x {log f2()
x f ( x )} = 0
Ï f1() x > 0, f1() x π1
fi f 2
(x) = f(x), Ì
Ó f2() x > 0, f2() x π1
Type 3: A logarithmic equation is of the form
log a
f 1
(x) = log a
f 2
(x), a > 0, a π 1
fi f 1
(x) = f 2
(x), f 1
(x) > 0 or f 2
(x) > 0
Type 4: A logarithmic equation is of the form
log
A=
log
f1() x f2()
x
Ï f1() x > 0, f1() x π1
Ô
fi f 1
(x) = f 2
(x), Ì or
Ô Ó f2() x > 0, f2() x π 1
Type 5: A logarithmic equation is of the form
log f(x)
g 1
(x) = log f(x)
g 2
(x)
Ï g1() x > 0, f() x > 0, π1
Ô
fi g 1
(x) = g 2
(x), Ì or
Ô Óg2() x > 0, f() x > 0, π 1
Type 6: A logarithmic equation is of the form
A
log f( x) = log f( x)
g1() x g2()
x
Ï g1() x > 0, π 1, f() x > 0
fi g 1
(x) = g 2
(x),
Ô
Ì or
Ô Óg2() x > 0, π 1, f() x > 0
Type 7: A logarithmic equation is of the form
2n log a
f 1
(x) = log a
f 2
(x), a > 0, a π 1, n Œ N
2n+1
fi f 1
(x) = f 2
(x), f 1
(x) > 0,
Type 8: A logarithmic equation is of the form
(2n + 1)log a
f 1
(x) = log a
f 2
(x), a > 0, a π 1, n Œ N
(2n+1)
fi f 1
(x) = f 2
(x), f 1
(x) > 0
Type 9: A logarithmic equation is of the form
log a
f(x) + log a
g(x) = log a
m(x), a > 0, a π 1
Ï f() x > 0
Ô
fi Ì gx () > 0
Ô
Ó f() x g() x = m()
x
Type 10: A logarithmic equation is of the form
log a
f(x) – log a
g(x) = log a
h(x) – log a
t(x)
where a > 0, a π 1
fi log a
f(x) + log a
t(x) = log a
g(x) + log a
h(x)
Ï f() x > 0,() t x > 0, g() x > 0, h() x > 0
fi Ì
Ó f() x ◊ t() x = g() x ◊h()
x
4. LOGARITHMIC IN-EQUATION
Type I: A logarithmic in-equation is of the form
Ï Ï gx () > 0
Ôlog a f( x) > log ag( x)
Ô
Ì
fi Ì a > 1
Ôa
> 1
Ô
Ó
Ó f() x > g()
x
Type II: A logarithmic in-equation is of the form
Ï Ï f() x > 0
Ôlog a f( x) > log ag( x)
Ô
Ì
fi Ì 0< a < 1
Ô0< a < 1
Ô
Ó
Ó f() x < g()
x
Type III: A logarithmic in-equation is of the form
Ï loga 0 0
1.
x > Ï x >
Ì fi Ì
ÔÓ
a > 1 Óa
> 1
2.
3.
4.
Ï loga x > 0 Ï0< x < 1
Ì fi Ì
ÔÓ
0 < a < 1 Ó0 < a < 1
Ï loga x < 0 Ï0< x < 1
Ì fi Ì
ÔÓ
a > 1 Ó a > 1
Ï loga x < 0 Ï x > 1
Ì fi Ì
ÔÓ
0 < a < 1 Ó0 < a < 1
LOGARITHM
CONCEPTUAL PROBLEM
EXAMPLE 1: Find the value of log 2
32.
Solution: We have
log 2
(32) = log 2
(2 5 )
= 5 ¥ log 2
(2)
= 5
Ê 1 ˆ
EXAMPLE 2: Find the value of log 3 Á .
Ë
˜
243¯
Solution: We have
Ê 1 ˆ
log3
Á
Ë
˜
243¯ = log 3 (3–5 )
= –5 ¥ log 3
(3)
= –5
EXAMPLE 3: Find the value of log (5).
5 5
Solution: We have
log (5) = log
5 5 5 3/2 (5)
3 3
= log 5(5)
=
2 2
EXAMPLE 4: Find the value of log 8
64.
Solution: We have
3
log 8
64 = log 3 (4 )
2