Luận văn Phương pháp giải phương trình, bất phương trình chứa logarit và các bài toàn liên quan

f(x) = log_ax, 0 < a = 1

D = (0; +∞)

0 < a = 1

f(x) = log_ax

a

I = R

x > 0

f⁰(x) =

1

.

x ln a

f(x) = log_ax

a > 1

ln a > 0

1

f⁰(x) =

> 0, ∀x > 0.

f(x) = log_ax

x ln a

a > 1

f(1) = 0, f(a) = 1

lim log_ax = −∞; lim log_ax = +∞.

x→+∞

x→0⁺

x

0

1

a

+∞

1 %

y = log_ax

0%

−∞%

0 < a < 1

f⁰(x) < 0, ∀x ∈ D

0 < a < 1

f(x) = log_ax

x

0

+∞

1

a

+∞

&0

y = log_ax

& 1

&−∞

f(x) = log_ax

0 < a < 1.

D = R⁺

a > 1

a > 0 a = 1

x₁

x₁, x₂∈ (0; +∞)

log_a(x₁x₂) = log_ax₁+ log_ax₂log

= log_ax₁− log_ax₂.

^ax₂

a > 0 a = 1

x > 0

α

log_ax^α= αlog_ax.

0 < a = 1, 0 < c = 1

log_cx

x > 0

log_ax =

.

log_cx

f(x) = log_ax (0 < a = 1)

1

x ∈ (0; +∞) (log_ax)⁰=

.

u = u(x)

x ln a

J ∈ R

y = log_au(x) (0 < a = 1)

u⁰(x)

(log_au(x))⁰=

.

u(x) ln a

a > 0 a = 1

x₁, x₂∈ (0; +∞)

•

a > 1

log_ax₁< log_ax₂⇔ x₁< x₂.

log_ax₁< log_ax₂⇔ x₁> x₂.

0 < a < 1

y = f(x)

f(c) = 0

[a; b]

f(a).f(b) < 0

c ∈ (a; b)

y = f(x)

[a; b] f(a) = A, f(b) = B

[a; b]

y = f(x)

f : [a; b] → R

(a; b) f(a) = f(b)

[a; b]

c ∈ (a; b)

f⁰(c) = 0

f(x)

[a; b]

f(x)

f(a) = f(b)

f(x)

[a; b]

M = m

f⁰(c) = 0

c ∈ (a; b)

f(c) = m

M > m

c ∈ (a; b)

f⁰(c) = 0

f(c) = M

f(x)

(a; b)

f⁰(x)

f(x)

(a; b)

n − 1

(a; b)

f(x)

(a; b)

f⁰(x)

(a; b)

f(x)

n + 1

(a; b)

f : [a; b] → R

(a; b)

∃c ∈ (a; b) :

[a; b]

f(b) − f(a)

b − a

f⁰(c) =

.

f(b) − f(a)

b − a

F(x) = f(x) −

[a; b]

x.

F(x)

(a; b)

F(a) = F(b)

c ∈ (a; b)

F⁰(c) = 0

f(b) − f(a)

F⁰(x) = f⁰(x) −

f⁰(c) =

b − a

(a; b)

F⁰(x) = 0

F(x)

f(x)

(a; b)

f⁰(x) > 0, ∀x ∈ (a; b)

f(x)

(a; b)

f⁰(x) < 0, ∀x ∈ (a; b)

a₁, a₂, ..., a_n

b₁, b₂, ..., b_n

(a₁b₁+ a₂b₂+ ... + a_nb_n)²≤ (a²₁+ a₂²+ ... + a_n²)(b²₁+ b₂²+ ... + b²_n).

∃k

a_i= kb_i, ∀i ∈ (1, 2, ..., n).

f(x) = (a²₁+a²₂+...+a_n²)x²−2(a₁b₁+a₂b₂+...+a_nb_n)x+(b²₁+b₂²+...+b_n²)

a²₁+ a²₂+ ... + a_n²= 0 ⇔ a₁= a₂= ... = a_n= 0

a²₁+ a²₂+ ... + a_n²> 0

f(x)

f(x) = (a₁x − b₁)²+ (a₂x − b₂)²+ ... + (a_nx − b_n)²≥ 0, ∀x ∈ R.

∆⁰= (a₁b₁+ a₂b₂+ ... + a_nb_n)²− (a₁²+ a²₂+ ... + a_n²)(b²₁+ b²₂+ ... + b_n²) ≤ 0

⇔ (a₁b₁+ a₂b₂+ ... + a_nb_n)²≤ (a²₁+ a₂²+ ... + a_n²)(b²₁+ b₂²+ ... + b²_n).



a x − b = 0





1

a₂x − b₂= 0

..................

a_nx − b_n= 0





∃k

a_i= kb_i, ∀i ∈ (1, 2, ..., n).

x > −1

ꢀ

(1 + x)^α≤ 1 + αx

0 ≤ α ≤ 1

(1 + x)^α≥ 1 + αx

α ≤ 0 ∨ α ≥ 1.

α = 0

α = 1

α < 0

α > 1

f(x) = (1 + x)^α− αx − 1

x > −1

h

i

f⁰(x) = α(1 + x)^α−1− α = α (1 + x)^α−1− 1

f⁰(x) = 0 ⇔ x = 0

f(x) ≥ 0, ∀x > −1 ⇔ (1 + x)^α≥ 1 + αx,

∀x > −1.

0 < α < 1

f(x)

f(x) ≤ 0, ∀x > −1 ⇔ (1 + x)^α≤ 1 + αx, ∀x > −1.

x

x − 1

ꢀ

x^α+ (1 − x)α ≤ 1

0 ≤ α ≤ 1

x^α+ (1 − x)α ≥ 1

α ≤ 0 ∨ α ≥ 1.

m > 0

a, b, c

a

b

c

+

= 0

m + 2 m + 1

m

ax²+ bx + c = 0

(0; 1)

a

b

c

F(x) =

x^m+2

+

x

^m+1+ x^m

m + 2

[0; 1]

m + 1

m

(0; 1)

F⁰(x) = x^m−1(ax²+ bx + c).

F(0) = F(1) = 0.

∃α ∈ (0; 1)

F⁰(α) = 0

⇔ α^m−1(aα²+ bα + c) = 0

⇔ aα²+ bα + c = 0.

ax²+ bx + c = 0

α ∈ (0; 1)

1

2014

1

< ln

<

.

2014

2013 2013

f(x) = ln x

1

f⁰(x) = .

x

f(x)

[2013; 2014]

(2013; 2014)

c ∈ (2013; 2014)

= f⁰(c) ⇔ ln 2014 − ln 2013 =

f(2014) − f(2013)

2014 − 2013

1

c

2014

2013

1

= .

c

⇔ ln

c ∈ (2013; 2014)

1

< <

c

1

.

2014

2013

1

2014

1

< ln

<

.

2014

2013 2013

f(x)

a (a ∈/ {0, 1, −1})

M ⊂ D(f)

ꢀ

∀x ∈ M ⇒ a^±1x ∈ M,

f(ax) = f(x), ∀x ∈ M.

f(x) = sin(2πlog₂x)

R⁺

f(x)

∀x ∈ R⁺

2^±1x ∈ R⁺

f(2x) = sin(2πlog₂(2x))

= sin(2π(1 + log₂x))

= sin(2πlog₂x) = f(x), ∀x ∈ R⁺.

f(5x) = f(x), ∀x > 0.

∀x ∈ R⁺⇒ 5^±x ∈ R⁺

log₅(5x) = 1 + log₅x ⇔ 2πlog₅(5x) = 2π + log₅x.

f(x) = cos[2πlog₅x], ∀x > 0,

f(5x) = cos[2πlog₅(5x)]

= cos[2π+2πlog₅x]

= cos[2πlog₅x] = f(x)

f(x) = cos(2πlog₅x), ∀x > 0

R⁺

f(x)

a (a ∈/ {0, 1, −1})

M

M ⊂ D(f)

ꢀ

∀x ∈ M ⇒ a^±1x ∈ M,

f(ax) = −f(x), ∀x ∈ M.

f(3x) = −f(x), ∀x > 0.

∀x ∈ R⁺⇒ 3^±x ∈ R⁺

log₃(3x) = 1 + log₃x ⇔ πlog₃(3x) = π + πlog₃x

f(x) = cos[πlog₃x],∀x > 0

f(3x) = cos[πlog₃(3x)]

= cos[π + πlog₃x]

= −cos[πlog₃x] = −f(x).

f(x) = cos(πlog₃x),

R⁺

ꢀ

0 < a = 1

f(x) = a^b

ꢀ

• log_af(x) = b ⇔

.

0 < a = 1

f(x) = g(x)

• log_af(x) = log_ag(x) ⇔

.

log_x(x²+ 4x − 4) = 3





ꢁ

√

ꢀ

x > −2 + 8

√

x²+ 4x − 4 > 0

0 < x = 1

√

⇔

⇔ 8 − 2 < x = 1.

x < −2 − 8



0 < x = 1

3 = log_xx³

(2.1) ⇔ log_x(x²+ 4x − 4) = log_xx³⇔ x²+ 4x − 4 = x³

"

x = 1

x = 2

x = −2

⇔ x³− x²− 4x + 4 = 0 ⇔ (x − 1)(x²− 4) = 0 ⇔

x = −2

x = 1, x = 2

2log₄(2x²− x + 2m − 4m²) + log (x²+ mx − 2m²) = 0

1

2

x₁, x₂

x²₁+ x²₂> 1

1

log₄(2x²− x + 2m − 4m²) = log₂(2x²− x + 2m − 4m²)

2

log (x²+ mx − 2m²) = −log₂(x²+ mx − 2m²).

1

2

log₂(2x²− x + 2m − 4m²) − log₂(x²+ mx − 2m²) = 0

⇔ log₂(2x²− x + 2m − 4m²) = log₂(x²+ mx − 2m²)

ꢀ

x²+ mx − 2m²> 0

⇔

2x²− x + 2m − 4m²= x²+ mx − 2m²





x²+ mx − 2m²> 0

ꢀ

x²+ mx − 2m²> 0

ꢁ

⇔

x₁= 2m

.

x²− (m + 1)x + 2m − 2m²= 0



x₂= 1 − m

x²₁+ x²₂> 1

x₁, x₂





"

2

(2m) + m(2m) − 2m > 0

−1 < m < 0

(1 − m)²+ m(1 − m) − 2m²> 0 ⇔

(2m)²+ (1 − m)²> 1

.

2

5

1

2

< m <



t = log_ax

x > 0

log^k_ax = t^k,

1

log_xa =

t

0 < x = 1.

a^{log c}= c^{log a}

t = a^{log x}

t = x^{log a}

b

a^{log x}

b

t = log_bx

(x − 2)^{log [9(x−2)]}= 9(x − 2)³.

3

x − 2 > 0 ⇔ x > 2.

log₃[(x − 2)^{log [9(x−2)]}] = log₃[9(x − 2)³]

3

⇔ [log₃[9(x − 2)]].log₃(x − 2) = 2 + log₃(x − 2)³

⇔ [2 + log₃(x − 2)].log₃(x − 2) = 2 + 3log₃(x − 2).

t = log₃(x − 2).

ꢁ

t = −1

(2 + t)t = 2 + 3t ⇔ t²− t − 2 = 0 ⇔

.

t = 2

7

•

t = −1

t = 2

log₃(x − 2) = −1 ⇔ x = .

3

log₃(x − 2) = 2 ⇔ x = 11.

7

3

x =

x = 11.

1

log_a(ax).log_x(ax) = log_a₂( ), 0 < a = 1.

a

ꢀ

ax > 0

0 < x = 1

⇔ 0 < x = 1.

1

(log_aa + log_ax).(log_xa + log_xx) = − log_aa

2

1

+ 1) = − .

2

⇔ (1 + log_ax).(

log_ax

t = log_ax

"

1

2

1

t = −

(1 + t).( + 1) = − ⇔ 2t²+ 5t + 2 = 0 ⇔

.

t

2

t = −2

1

2

1

√

•

t = −

log_ax = − ⇔ x =

.

2

a

1

t = −2

log_ax = −2 ⇔ x =

.

a²

1

a²

√

x =

.

a

∆

lg²x − lg x.log₂(4x) + 2log₂x = 0.

x > 0

lg²x − (2 + log₂x) lg x + 2log₂x = 0.

t = lg x

t²− (2 + log₂x).t + 2log₂x = 0

∆ = (2 + log₂x)²− 8log₂x = (2 − log₂x)².

ꢁ

t = 2

.

t = log₂x

•

t = 2

lg x = 2 ⇔ x = 100.

lg x = log₂x ⇔ lg x =

lg x

t = log₂x

⇔ lg x = 0 ⇔ x = 1.

lg 2

x = 100, x = 1.

lg²(x²+ 1) + (x²− 5). lg(x²+ 1) − 5x²= 0.

t = lg(x²+1)

t ≥ 0 x²+1 ≥ 1

lg(x²+1) ≥ lg 1 = 0.

t²+ (x²− 5).t − 5x²= 0

∆ = (x²− 5)²+ 20x²= (x²+ 5)².

ꢁ

t = 5

.

t = −x²

√

•

t = 5

lg(x²+ 1) = 5 ⇔ x²+ 1 = 10⁵⇔ x = ± 99999.

ꢀ

lg(x²+ 1) = 0

t = −x²

lg(x²+ 1) = −x²⇔

⇔ x = 0.

x²= 0

√

x = ± 99999

x = 0.

•

q

3 + log₂(x²− 4x + 5) + 2 5 − log₂(x²− 4x + 5) = 6.





2

x − 4x + 5 > 0

3 + log₂(x²− 4x + 5) ≥ 0 ⇔ x²− 4x + 5 ≤ 2⁵⇔ x²− 4x − 27 ≤ 0



5 − log₂(x²− 4x + 5) ≥ 0

√

⇔ 2 − 29 ≤ x ≤ 2 + 29.

ꢀ

p

u = 3 + log₂(x²− 4x + 5)

p

,

u, v ≥ 0

v = 5 − log₂(x²− 4x + 5)