Oblicz: [latex]log _{ frac{1}{4} } sqrt{2} [/latex]

log a b= c ---> a^c=b więc: log 1/4 v2 =x 1/4^x=v2 (4^-1)^x=2^1/2 (2^-2)^x=2^1/2 2^-2x=2^1/2 opuszczam podstawy: -2x=1/2 /*(-1/2) x=-1/4

[latex]log_{frac{1}{4}}sqrt{2}=log_{2^{-2}}2^{frac{1}{2}}=x[/latex] [latex](2^{-2})^x=2^{frac{1}{2}}[/latex] [latex]-2x=frac{1}{2};/:(-2)[/latex] [latex]x=-frac{1}{4}[/latex] Stąd masz [latex]log_{frac{1}{4}}sqrt{2}=-frac{1}{4}[/latex]

1. Oblicz: a) [latex]log_{ { sqrt{5} } }5 sqrt{5}[/latex] = b) [latex]log x_{ sqrt[3]{3} } 27[/latex] = c) [latex]log_{2}8 sqrt{2} [/latex] = d) [latex]log_{ frac{1}{3} }81sqrt{3} [/latex] =

1. Oblicz: a) [latex]log_{ { sqrt{5} } }5 sqrt{5}[/latex] = b) [latex]log x_{ sqrt[3]{3} } 27[/latex] = c) [latex]log_{2}8 sqrt{2} [/latex] = d) [latex]log_{ frac{1}{3} }81sqrt{3} [/latex] =...

Pilne.Oblicz. a.) [latex]log_{frac{1}{5}} 5sqrt{5}= [/latex] b.) [latex]log_{frac{4}{3}}(log_{4}2sqrt{2})=[/latex] c.) [latex]log_{36}6+log_{16}4-log_{8}2=[/latex] d.) [latex]frac{25}{5frac{3}{2}} =[/latex] e.) [latex]frac{3frac{3}{2}}{sqrt{3}}=[/latex]

Pilne.Oblicz. a.) [latex]log_{frac{1}{5}} 5sqrt{5}= [/latex] b.) [latex]log_{frac{4}{3}}(log_{4}2sqrt{2})=[/latex] c.) [latex]log_{36}6+log_{16}4-log_{8}2=[/latex] d.) [latex]frac{25}{5frac{3}{2}} =[/latex] e.) [latex]frac{3frac{3}{2}}{sqrt{3}}=[/la...

1. Oblicz: 1) [latex] log_{ sqrt{2} } 2 sqrt{2} [/latex] 2. Oblicz podstawę logarytmu:  1) [latex]log_{a}2 = frac{1}{2} [/latex] 2) [latex] log_{a}8=6 [/latex] 3)[latex] log_{a}32=-5 [/latex] Proszę o pomoc :P

1. Oblicz: 1) [latex] log_{ sqrt{2} } 2 sqrt{2} [/latex] 2. Oblicz podstawę logarytmu:  1) [latex]log_{a}2 = frac{1}{2} [/latex] 2) [latex] log_{a}8=6 [/latex] 3)[latex] log_{a}32=-5 [/latex] Proszę o pomoc :P...

Oblicz: 8 [latex] log_{2} frac{1}{ sqrt{2} } [/latex]

Oblicz: 8 [latex] log_{2} frac{1}{ sqrt{2} } [/latex]...

Oblicz: a) [latex]log frac{1}{3} 81 sqrt[4]{3}= [/latex] b) Wiedząc, że [latex]log 5^{2}= a [/latex] oblicz [latex]log _{5} frac{2}{25} [/latex] c) [latex][ (frac{2}{3}) ^{-1}-4 ^{-1}] ^{-1}= [/latex] d) [latex]( sqrt{3}- sqrt{2}) ^{2}= [/latex]

Oblicz: a) [latex]log frac{1}{3} 81 sqrt[4]{3}= [/latex] b) Wiedząc, że [latex]log 5^{2}= a [/latex] oblicz [latex]log _{5} frac{2}{25} [/latex] c) [latex][ (frac{2}{3}) ^{-1}-4 ^{-1}] ^{-1}= [/latex] d) [latex]( sqrt{3}- sqrt{2}) ^{2}= ...