Question

3. Podaj dziedzinę wyrażenia, a następnie przedstaw je w postaci logarytmu pewnej liczby.
Podpunkty f) g) h) ​​

Answer 1

Odpowiedź:

zad. 3

[tex]f)~~\dfrac{1}{3} log_{5}8x^{3}-2log_{5}y\sqrt{x} +\dfrac{1}{2} =log_{5}\dfrac{2\sqrt{5} }{y^{2}} ,~zal.~~x > 0,~~y > 0[/tex]

[tex]g)~~\dfrac{1}{2} log_{2}x-log_{2}y-2=log_{2}\dfrac{\sqrt{x} }{4y} ,~~zal.~~x > 0,~~y > 0[/tex]

[tex]h)~~\dfrac{1}{2} log4x^{4}+\dfrac{1}{3} log8x^{6}+\dfrac{1}{4} log16x^{8}-3=log\dfrac{x^{6}}{125} ,~~zal.~~x > 0[/tex]

Szczegółowe wyjaśnienie:

                       Logarytmy

Logarytmem liczby b przy podstawie a nazywamy taką liczbę c, że  a podniesione do potęgi  c daje liczbę b.

          [tex]\boxed{~~log_{a}b=c~~\Rightarrow ~~a^{c}=b,~~zal.~a\neq 1,~~a > 0,~~b > 0~~}[/tex]

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

Rozwiązanie:

zad. 3

[tex]f)\\\\zal.~~x > 0,~~y > 0\\\\\dfrac{1}{3} log_{5}8x^{3}-2log_{5}y\sqrt{x} +\dfrac{1}{2} =log_{5}(8x^{3})^{\frac{1}{3} }-log_{5}(y\sqrt{x} )^{2}+log_{5}5^{\frac{1}{2} }=log_{5}\sqrt[3]{8x^{3}} -log_{5}y^{2}(\sqrt{x} )^{2} +log_{5}\sqrt{5} =log_{5}2x-log_{5}y^{2}x+log_{5}\sqrt{5} =log_{5}\dfrac{2x}{y^{2}x}+log_{5}\sqrt{5} =log_{5}\dfrac{2}{y^{2}} +log_{5}\sqrt{5} =log_{5}\dfrac{2\sqrt{5} }{y^{2}}[/tex]

[tex]g)\\\\zal.~x > 0,~y > 0\\\\\dfrac{1}{2} log_{2}x-log_{2}y-2=log_{2}(x)^{\frac{1}{2} } -log_{2}y-log_{2}2^2=log_{2}\sqrt{x} -log_{2}y-log_{2}4=log_{2}\dfrac{\sqrt{x} }{y} -log_{2}4=log_{2}\dfrac{\sqrt{x} }{4y}[/tex]

[tex]h)\\\\zal.~~x > 0\\\\\dfrac{1}{2} log4x^{4}+\dfrac{1}{3} log8x^{6}+\dfrac{1}{4} log16x^{8}-3=log\sqrt[2]{4x^{4}} +log\sqrt[3]{8x^{6}} +log\sqrt[4]{16x^{8}} -log10^{3}=log(2x^{2}) +log(2x^{2} )+log(2x^{2} )-log1000=log(2x^{2} \cdot 2x^{2} \cdot 2x^{2} )-log1000=log(8x^{6})-log1000=log\dfrac{8x^{6}}{1000} =log\dfrac{x^{6}}{125}[/tex]