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cum rezolv si eu ecuatia asta exponentiala ? 0,25 la puterea 5-x = 2 la puterea x+3 

Răspuns :

JohnAndrew
[tex](\frac{1}4} )^{5-x} = 2^{x+3}[/tex]

-logaritmezi in baza e;
-exponentul iese in fata;

[tex](5-x)*ln(1/4)=(x+3)*ln(2)[/tex]

*ln(a/b)=ln a - ln b;

[tex](5-x)ln(1)-(5-x)ln2^{2} =(x+3)*ln(2)[/tex]

ln(1)=0;

[tex]-2(5-x)ln(2)=(x+3)ln(2)[/tex]  

-impartim cu ln 2

[tex]-10+2x=x+3[/tex]
[tex]x=13[/tex]

verificare:

[tex](\frac{1}{4} )^{-8} =2^{16} [/tex]

[tex]({2^{-2}})^{-8}=2^{16} [/tex]

 [tex]2^{16}=2^{16} [/tex]
Karinaszutor
(\frac{1}4} )^{5-x} = 2^{x+3} -logaritmezi in baza e; -exponentul iese in fata; (5-x)*ln(1/4)=(x+3)*ln(2) *ln(a/b)=ln a - ln b; (5-x)ln(1)-(5-x)ln2^{2} =(x+3)*ln(2) ln(1)=0; -2(5-x)ln(2)=(x+3)ln(2) -impartim cu ln 2 -10+2x=x+3 x=13 verificare: (\frac{1}{4} )^{-8} =2^{16} ({2^{-2}})^{-8}=2^{16} 2^{16}=2^{16}

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