Question

SZYBKO BŁAGAM DAJE NAJ

Answer 1

Odpowiedź:

[tex]\huge\boxed{\huge\boxed{~~zad.~3~~}}[/tex]

[tex]\huge\boxed{~~a)~~log_{3}9=2~~}[/tex]   [tex]\huge\boxed{~~b)~~log175-log1\frac{3}{4}=2~~}[/tex]

[tex]\huge\boxed{~~c)~~log_{3}6+log_{3}1,5=2~~}[/tex]

[tex]\huge\boxed{~~d)~~log_{2}\sqrt{18} -log_{2}3=\dfrac{1}{2} ~~}[/tex]

Szczegółowe wyjaśnienie:

Korzystamy ze wzorów:

• [tex]log_{a}b^{n}=n\cdot log_{a}b,~~zal.~~a\neq 1,a > 0,b > 0[/tex]
• [tex]log_{a}b+log_{a}c=log_{a}(b\cdot c),~~zal.~~a\neq 1,a > 0,b > 0,c > 0[/tex]
• [tex]log_{a}b-log_{a}c=log_{a}\frac{b}{c} ,~~zal.~~a\neq 1,a > 0,b > 0,c > 0[/tex]
• [tex]log_{a}a=1,~~zal.~~a\neq 1,a > 0[/tex]
• [tex]logb=log_{10}b,~~zal.~~b > 0[/tex]
• [tex]\sqrt[n]{x^{n}} =x^{n\cdot \frac{1}{n} } =x,~~zal.~~x\geq 0[/tex]

Obliczamy:

[tex]\huge\boxed{~~zad.~3~~}[/tex]

[tex]\huge\boxed{a)}~~log_{3}9=log_{3}3^{2}=2\cdot log_{3}3=2\cdot 1=\boxed{2}[/tex]

[tex]\huge\boxed{b)}~~log175-log1\frac{3}{4} =log175-log\frac{7}{4} =log(175\div \frac{7}{4} )=log(\frac{175\!\!\!\!\!\diagup^2^5}{1} \cdot \frac{4}{7\!\!\!\!\diagup_1} )=log100=log_{10}100=log_{10}10^{2}=2\cdot log_{10}10=2\cdot 1=\boxed{2}[/tex]

[tex]\huge\boxed{c)}~~log_{3}6+log_{3}1,5=log_{3}6+log_{3}\frac{3}{2} =log_{3}(6\!\!\!\diagup^3\cdot \frac{3}{2\!\!\!\!\diagup_1})=log_{3}9=log_{3}3^{2}=2\cdot log_{3}3=2\cdot 1=\boxed{2}[/tex]

[tex]\huge\boxed{d)}~~log_{2}\sqrt{18} -log_{2}3=log_{2}\sqrt{9\cdot 2} -log_{2}3=log_{2}\sqrt{3^{2}\cdot 2} -log_{2}3=log_{2}3\sqrt{2} -log_{2}3=log_{2}\frac{3\!\!\!\!\diagup^1\sqrt{2} }{3\!\!\!\!\diagup_1} =log_{2}\sqrt{2} =log_{2}2^{\frac{1}{2} }=\dfrac{1}{2} \cdot log_{2}2=\dfrac{1}{2} \cdot 1=\boxed{\dfrac{1}{2}}[/tex]