Potencije

Uvod

Ponovimo

Prisjetimo se da umnožak [latex]5 · 5[/latex]

u kojemu se faktor [latex]\textcolor{blue}5[/latex] pojavljuje dva puta 

kraće zapisujemo [latex]\textcolor{blue}5^\textcolor{red}2[/latex]. 

 Umnožak [latex]10 ·10  ·10  ·10  ·10[/latex] u kojemu se

pet puta pojavljuje broj [latex]\textcolor{blue}{10}[/latex] možemo 

kraće zapisati [latex]\textcolor{blue}{10}^\textcolor{red}5[/latex], što čitamo: deset na petu.

Na isti način označava se umnožak bilo kojih jednakih brojeva. 

Tako je: [latex]\bm{a · a · a · a · a = a^5}[/latex],

što čitamo: [latex]\bm{a}[/latex] na petu.

Nauči

Umnožak [latex]n[/latex] jednakih brojeva [latex]a[/latex] naziva se [latex]n[/latex]–ta POTENCIJA broja [latex]a[/latex] i označava [latex]a^n[/latex].
[latex]a^n[/latex] čitamo: [latex]a[/latex] na entu ili [latex]a[/latex] na en.

[latex]\underbrace{a \cdot a\cdot a\cdot... \cdot a}=a^n \\ n \enspace faktora[/latex]

[latex]\textcolor{blue}a[/latex] je baza potencije [latex]\textcolor{blue}a^\textcolor{red}n[/latex].

[latex]\textcolor{red}n[/latex] je ekponent potencije [latex]\textcolor{blue}a^\textcolor{red}n[/latex]. 

Primjer 1.

Broj [latex]\textcolor{blue}2[/latex] je baza, a [latex]\textcolor{red}6[/latex] je eksponent.

Broj [latex]\textcolor{green}{2^6}[/latex] je potencija.

[latex]\textcolor{blue}{2}[/latex] [latex]· \textcolor{blue}{2}[/latex] [latex]· \textcolor{blue}{2}[/latex] [latex]· \textcolor{blue}{2}[/latex] [latex] · \textcolor{blue}{2}[/latex][latex] ·  \textcolor{blue}{2}[/latex] [latex] = \textcolor{blue}{2}[/latex][latex]^\textcolor{red}6[/latex]

Istaknimo da vrijedi:

[latex]\pmb{a^1=a}[/latex]

npr. [latex]\pmb{2^1=2}[/latex]

Zadatak 1.

Zadatak 2.

Primjer 2.

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2  = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

[latex](–2)^4 = (–2)·(–2)·(–2)·(–2) = 16[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

[latex](–2)^4 = (–2)·(–2)·(–2)·(–2) = 16[/latex]

[latex](–2)^5 = (–2)·(–2)·(–2)·(–2)·(–2) = –32[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

[latex](–2)^4 = (–2)·(–2)·(–2)·(–2) = 16[/latex]

[latex](–2)^5 = (–2)·(–2)·(–2)·(–2)·(–2) = –32[/latex]

[latex](–2)^6 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2) = 64[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

[latex](–2)^4 = (–2)·(–2)·(–2)·(–2) = 16[/latex]

[latex](–2)^5 = (–2)·(–2)·(–2)·(–2)·(–2) = –32[/latex]

[latex](–2)^6 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2) = 64[/latex]

[latex](–2)^7 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2) = –128[/latex]

 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex](–2)^1 = –2[/latex]

[latex](–2)^2 = (–2)·(–2) = 4 [/latex]

[latex](–2)^3 = (–2)·(–2)·(–2) = –8[/latex]

[latex](–2)^4 = (–2)·(–2)·(–2)·(–2) = 16[/latex]

[latex](–2)^5 = (–2)·(–2)·(–2)·(–2)·(–2) = –32[/latex]

[latex](–2)^6 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2) = 64[/latex]

[latex](–2)^7 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2) = –128[/latex]

[latex](–2)^8 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2)= 252[/latex]

Što zaključuješ? 

Promotrimo potencije s negativnom bazom, primjerice baza neka bude [latex](–2)[/latex]. 

[latex]\textcolor{blue}{(–2)^1 = –2}[/latex]

[latex]\textcolor{red}{(–2)^2 = (–2)·(–2) = 4}[/latex]

[latex]\textcolor{blue}{(–2)^3 = (–2)·(–2)·(–2) = –8}[/latex]

[latex]\textcolor{red}{(–2)^4 = (–2)·(–2)·(–2)·(–2) = 16}[/latex]

[latex]\textcolor{blue}{(–2)^5 = (–2)·(–2)·(–2)·(–2)·(–2) = –32}[/latex]

[latex]\textcolor{red}{(–2)^6 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2) = 64}[/latex]

[latex]\textcolor{blue}{(–2)^7 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2) = –128}[/latex]

[latex]\textcolor{red}{(–2)^8 = (–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2)·(–2)= 252}[/latex]

Što zaključuješ? 

Potencija s negativnom bazom i parnim eksponentom ima pozitivnu vrijednost.

Potencija s negativnom bazom i neparnim eksponentom ima negativnu vrijednost.

Zadatak 3.

Primjer 3.

Izračunaj vrijednost izraza [latex]3a^3+2a^4-5[/latex] ako je: 

a/ [latex]a=3[/latex]
b/ [latex]a=-2[/latex]

$$3\red{a}^3+2\red{a}^4-5$$

 

 

 

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot\red{3}^3+2\cdot\red{3}^4-5$$

 

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot\red{3}^3+2\cdot\red{3}^4-5$$

[latex]=3\cdot27+2\cdot81-5[/latex]

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot\red{3}^3+2\cdot\red{3}^4-5$$

[latex]=3\cdot27+2\cdot81-5[/latex]

[latex]=81+162-5[/latex]

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot\red{3}^3+2\cdot\red{3}^4-5$$

[latex]=3\cdot27+2\cdot81-5[/latex]

[latex]=81+162-5[/latex]

[latex]=238[/latex]

$$3\red{a}^3+2\red{a}^4-5$$

 

 

 

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot(\red{-2})^3+2\cdot(\red{-2})^4-5$$

 

 

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot(\red{-2})^3+2\cdot(\red{-2})^4-5$$

[latex]=3\cdot(-8)+2\cdot16-5[/latex]

 

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot(\red{-2})^3+2\cdot(\red{-2})^4-5$$

[latex]=3\cdot(-8)+2\cdot16-5[/latex]

[latex]=-24+32-5[/latex]

 

$$3\red{a}^3+2\red{a}^4-5$$

$$3\cdot(\red{-2})^3+2\cdot(\red{-2})^4-5$$

[latex]=3\cdot(-8)+2\cdot16-5[/latex]

[latex]=-24+32-5[/latex]

[latex]=3[/latex]

Zadatak 4.

Zadatak 5.

Pokušajmo riješiti zadatak iz uvoda, odnosno uz pomoć potencija brzo doći do broja zrna na šahovskoj ploči.