2.3 Eksponentielle ulikheter
Oppgave 2.30
a) 5 5 5 lg 3 0 lg52 lg 5 lg 2
2 3 5 3 lg 3 lg lg 3 lg
2 2 2 lg 3 lg 3
x x x
x x x
> −
⋅ < ⇔ < ⇔ < ⇔ ⋅ < ⇔ < ⇔ <
b)
( ) ( )
( )
Utvider brøken med -1 lg 2 lg 3 0
5 2 2 3 lg 5 2 lg 2 3 lg 5 lg 2 lg 2 lg 3
lg 5 lg 2 lg 2 lg 3 lg 2 lg 3 lg 2 lg 5
lg 2 lg 5 lg 5 lg 2
lg 2 lg 3 lg 2 lg 5
lg 2 lg 3 lg 3 lg 2
x x x x x x
x x x x
x x x
− <
⋅ ≥ ⋅ ⇔ ⋅ ≥ ⋅ ⇔ + ≥ + ⇔
+ ⋅ ≥ + ⋅ ⇔ ⋅ − ⋅ ≥ − ⇔
− −
⋅ − ≥ − ⇔ ≤
≤ ⇔ −
−
Oppgave 2.31
a) lg1,04 0 lg1, 5
1, 04 1, 5 lg1, 04 lg1, 5 lg1, 04 lg1, 5 10, 3
lg1, 04
x x
x x x
> ⇔ > ⇔ ⋅ > ⇔> > ⇔ >
b)
( )
( )
lg1,04 lg1,02 01200 1, 02
1000 1, 04 1200 1, 02 1, 04 lg1, 04 lg 1, 2 1, 02
1000
lg1, 04 lg1, 2 lg1, 02 lg1, 04 lg1, 2 lg1, 02 lg1, 04 lg1, 02 lg1, 2 lg1, 04 lg1, 02 lg1, 2
lg1, 2 lg1, 04 lg1, 02 9, 4
x
x x x x x
x x
x x
x x x
x x
− >
⋅ ≤ ⋅ ⇔ ≤ ⋅ ⇔ ≤ ⋅ ⇔
≤ + ⇔ ⋅ ≤ + ⋅ ⇔
⋅ − ⋅ ≤ ⇔ ⋅ − ≤ ⇔
≤ ⇔ ≤
−
Oppgave 2.32 a)
lg1,03 0
12000 ( ) 12000 10000 1, 03 12000 1, 03
10000 lg1, 2
lg1, 03 lg1, 2 lg1, 03 lg1, 2 6,17
lg1, 03
Når beløpet har stått i banken mer enn 6,17 år (ca. 6 år og 2 mnd) har det vokst til over 12000kr
x x
x
A x
x x x
>
> ⇒ ⋅ > ⇔ > ⇔
> ⇔ ⋅ > ⇔ > ⇔ >
b)
( )
Med 2% rente har Bjørns 12000kr etter år vokst til ( ) 12000 1, 02
12000 1, 02 ( ) ( ) 10000 1, 03 12000 1, 02 1, 03
10000
1, 03 1, 2 1, 02 lg1, 03 lg 1, 2 1, 02 lg1, 03 lg1, 2 lg1, 02 lg1, 03 lg1, 2
x
x
x x x
x x x x x x
x B x
A x B x
x x
= ⋅
> ⇒ ⋅ > ⋅ ⇔ > ⋅ ⇔
> ⋅ ⇔ > ⋅ ⇔ > + ⇔
⋅ > +
( )
lg1,03 lg1,02 0lg1, 02 lg1, 03 lg1, 02 lg1, 2 lg1, 2
lg1, 03 lg1, 02 lg1, 2 18, 7
lg1, 03 lg1, 02
Anne har mer penger enn Bjørn i banken etter mer enn 18,7 år (ca. 18 år og 8mnd)
x x
x x x
− >
⋅ ⇔ ⋅ − ⋅ > ⇔
⋅ − > ⇔ > ⇔ >
−
Oppgave 2.33 a)
( )
( ) ( )
( )
1 3
1 1
1 3 1
3
3 2
0 3
Nullpunkt for teller: 3 2 0 3 2 lg 3 lg 2 lg 3 lg 2 lg 2 0, 6 lg 3 3 er voksende, derfor er 3 2 0 når lg 2
lg 3
Nullpunkt for nevner: 3 0 3 3 3 3 1 1
er minkende, derfor e
x x
x x x
x x
x x x
x
x x
x
x x
− −
− >
−
− = ⇔ = ⇔ = ⇔ ⋅ = ⇔ = ≈
− > >
− = ⇔ = ⇔ = ⇔ − = ⇔ = −
( )
13r 3x− >0 når x< −1
( )
133 2 lg 2
0 når 1 3 lg 3
x
x− x
> − < <
−
–3 –2 –1 0 lg 2lg 3 1 2
x 3x - 2
( )
13 x−3( )
133 2
3
x x
−
−
b)
2
2 4
3 2 6 0
Nullpunkt for teller: 2 4 0 2 4 2 2 2
2 er voksende, derfor er 2 4 0 når 2
Nullpunkt for nevner: 3 2 6 0 2 2 1
2 er voksende, derfor er 3 2 6 0 når 1
x x
x x x
x x
x x
x x
x x
x x
− >
⋅ −
− = ⇔ = ⇔ = ⇔ =
− > >
⋅ − = ⇔ = ⇔ =
⋅ − > >
2 4
0 når 1 og når 2 3 2 6
x
x− > x< x>
⋅ − c)
4 3
2 2
2 8
2 2 0
2 8
2 2 (2 8)
2 8 2 8 0
2 2 2 16
2 8 0 (2 16)
2 8 0
Nullpunkt for teller: 2 16 0 2 16 2 2 4
Nullpunkt for nevner: 2 8 0 2 8 2 2 3
2 er voksende, derfor er 2
x x
x x
x x
x x
x x
x x x
x x x
x x x
x x
x x
− >
− >
−
⋅ −
− >
− −
− ⋅ +
− >
− − >
−
− = ⇔ = ⇔ = ⇔ =
− = ⇔ = ⇔ = ⇔ =
16 0 når 4 x og 2x 8 0 når 3.x
− > > − > >
2 2 når 3 4.
2 8
x
x > < <x
−
–1 0 1 2 3 4
x 2x – 16
2x−8 –1
(2 16)
2 8
x x
− −
−
–1 0 1 2 3 4
x 2x – 4
3 2⋅ −x 6
2 4
3 2 6
x x
−
⋅ −
Oppgave 2.34
a)
( ) ( ) ( )
lg 2 lg 3
Nullpunkter
3 3 3 2
2 2
Nullpunkt Nullpunkt 1
3 5 3 6 0 3 5 3 6 0 3 3 3 2 0
x x
x x x x x x
x x
= ∨ =
= =
− ⋅ + > ⇔ − ⋅ + > ⇔ − ⋅ − >
2 lg 2
3 5 3 6 0 når og når 1
lg 3
x x
x x
− ⋅ + > < >
b)
( ) ( ) ( )
lg 3 lg 2
Nullpunkter
2 3 2 1
2 2
Nullpunkt Ingen nullpunkt
2 2 2 3 2 2 2 3 0 2 3 2 1 0
x x
x x x x x x
x
= ∨ =−
=
− ⋅ > ⇔ − ⋅ − > ⇔ − ⋅ + >
2 lg 3
2 2 2 3 når
lg 2
x x
− ⋅ > x>
–3 –2 –1 0 1 lg 3lg 22
x 2x - 3
2x +1
22x− ⋅ −2 2x 3
–3 –2 –1 0 lg 2lg 3 1 2
x 3x - 3
3x - 2
32x− ⋅ +5 3x 6
c)
( )
( ) ( ) ( ) ( ) ( )
N
lg 3 lg 2
Nullpunkt Nullpunkter
Nullpunkt
2 1 og 2 3 0
2 2
Nullpunkt 1
2 2 3 2 2 2
2 2 3 2 2 3
2 0 2 0
2 2 2 2 2 2
2 2 3 2 2 2 2 4 2 3 1 2 1 2 3
0 0 0
2 2 2 2 2 2
x x
x x x
x x
x x
x x x
x x
x x x x x x x
x x x
x
= = = =
=
⋅ − − ⋅ −
⋅ − ⋅ −
> ⇔ > − ⇔ < ⇔
− − −
⋅ − − + ⋅ − + ⋅ − − ⋅ − ⋅ −
< ⇔ < ⇔ <
− − −
2 2 3 lg 3
2 når 0 1 og når
2 2 lg 2
x x
x x x
> ⋅ − < < >
−
–3 –2 –1 0 1 lg 3lg 2 2
x -1
2x - 1 2x -3 2x-2 Brøk
