Ako je [inlmath]\log_2 x = a[/inlmath], [inlmath]\log_4 y = b[/inlmath], onda je [inlmath]\log_8 \frac{1}{xy}[/inlmath] jednako:
[inlmath]A.\quad\frac{a+b}{b}[/inlmath]
[inlmath]B.\quad\frac{2a-b}{3}[/inlmath]
[inlmath]C.\quad -\frac{a}{2b}[/inlmath]
[inlmath]D.\quad -\frac{a+2b}{3}[/inlmath]
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Logaritam
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Re: Logaritam
od blake » Utorak, 07. Maj 2013, 17:18
Prvo primjeniš pravilo [inlmath]\log_ax=y\;\Leftrightarrow\;x=a^y[/inlmath]
Tj. [inlmath]x=2^a[/inlmath] i [inlmath]y=4^b[/inlmath]
[dispmath]\log_8(2^{-a}\cdot 2^{-2b})=\frac{\log(2^{-a-2b})}{\log 8}[/dispmath]
To je pravilo [inlmath]\log_ab=\frac{\log_cb}{\log_ca}[/inlmath], [inlmath]c=10[/inlmath] sam ja stavija jer desetku ne trebaš pisat
[dispmath]\frac{\log(2^{-a-2b})}{\log 8}=\frac{-a-2b\cdot\log 2}{\log 2^3}=\frac{-a-2b\cdot\log 2}{3\cdot\log 2}[/dispmath]
Izbaciš [inlmath]\log 2[/inlmath] iz brojnika i nazivnika i ostane ti:
[dispmath]\frac{-a-2b}{3}[/dispmath]
Tj. [inlmath]x=2^a[/inlmath] i [inlmath]y=4^b[/inlmath]
[dispmath]\log_8(2^{-a}\cdot 2^{-2b})=\frac{\log(2^{-a-2b})}{\log 8}[/dispmath]
To je pravilo [inlmath]\log_ab=\frac{\log_cb}{\log_ca}[/inlmath], [inlmath]c=10[/inlmath] sam ja stavija jer desetku ne trebaš pisat
[dispmath]\frac{\log(2^{-a-2b})}{\log 8}=\frac{-a-2b\cdot\log 2}{\log 2^3}=\frac{-a-2b\cdot\log 2}{3\cdot\log 2}[/dispmath]
Izbaciš [inlmath]\log 2[/inlmath] iz brojnika i nazivnika i ostane ti:
[dispmath]\frac{-a-2b}{3}[/dispmath]
Re: Logaritam
od forzajuve » Utorak, 07. Maj 2013, 23:20
Jedna mala opaska - nadam se da se kolega blake nece ljutiti - ja bih ovo
napisao ovako
[dispmath]\frac{(-a-2b)\cdot\log 2}{3\cdot\log 2}[/dispmath]
blake je napisao:[dispmath]\frac{-a-2b\cdot\log 2}{3\cdot\log 2}[/dispmath]
napisao ovako
[dispmath]\frac{(-a-2b)\cdot\log 2}{3\cdot\log 2}[/dispmath]
Re: Logaritam
od Daniel » Sreda, 08. Maj 2013, 00:03
I od mene – bravo! 
I meni promaklo.
I meni promaklo.
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
Re: Logaritam
od smAshh » Petak, 10. Maj 2013, 20:45
[inlmath]\log_a x=2[/inlmath]
[inlmath]\log_b x=3[/inlmath]
[inlmath]\log_c x=6[/inlmath]
onda je [inlmath]\log_{abc}x[/inlmath] jednako?
[inlmath]\log_b x=3[/inlmath]
[inlmath]\log_c x=6[/inlmath]
onda je [inlmath]\log_{abc}x[/inlmath] jednako?
Poslednji put menjao smAshh dana Subota, 11. Maj 2013, 12:22, izmenjena samo jedanput
-
smAshh
Re: Logaritam
od Daniel » Subota, 11. Maj 2013, 00:43
Ajd, molim te, napiši ove izraze malo preglednije, da ne bismo nagađali kako oni zapravo treba da glase.
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
Re: Logaritam
od Daniel » Subota, 11. Maj 2013, 12:35
E ovako je već bolje. [dispmath]\log_{abc}x=\frac{1}{\log_x abc}=\frac{1}{\log_x a+\log_x b+\log_x c}=\frac{1}{\frac{1}{\log_a x}+\frac{1}{\log_b x}+\frac{1}{\log_c x}}=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=1[/dispmath]
[dispmath]\log_{abc}x=\frac{1}{\log_x abc}=\frac{1}{\log_x a+\log_x b+\log_x c}=\frac{1}{\frac{1}{\log_a x}+\frac{1}{\log_b x}+\frac{1}{\log_c x}}=\frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=1[/dispmath]I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
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