Ответ:
[tex]\displaystyle \boldsymbol {z=5^{(-\displaystyle \frac{5}{12} )}}[/tex]
Объяснение:
Преобразуем выражение под логарифмом
[tex]\displaystyle \sqrt[3]{5*\sqrt[4]{5} } =\bigg(5*5^{(1/4)}\bigg)^{(1/3)}=\bigg(5^{(5/4)}\bigg)^{1/3)}=5^{5/12}[/tex]
Теперь мы легко найдем z.
[tex]\displaystyle log_z\bigg(5^{(5/12)}\bigg)=-1\\\\\\\frac{5}{12} log_z(5)=-1\qquad \bigg|\;log_a(b)=\frac{log_c(b)}{log_c(a)} \\\\\\\frac{lg(5)}{lg(z)} =-\frac{12}{5} \\\\\\lg(z) = -\frac{5}{12} lg(5)\\\\\\lg(z) = lg(5^{(-5/12)}})\\\\\\z=5^{(-5/12)}\\\\\\z=\frac{1}{5^{(5/12)}}\\\\\\z=5^{(- \displaystyle \frac{5}{12})[/tex]
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Answers & Comments
Ответ:
[tex]\displaystyle \boldsymbol {z=5^{(-\displaystyle \frac{5}{12} )}}[/tex]
Объяснение:
Преобразуем выражение под логарифмом
[tex]\displaystyle \sqrt[3]{5*\sqrt[4]{5} } =\bigg(5*5^{(1/4)}\bigg)^{(1/3)}=\bigg(5^{(5/4)}\bigg)^{1/3)}=5^{5/12}[/tex]
Теперь мы легко найдем z.
[tex]\displaystyle log_z\bigg(5^{(5/12)}\bigg)=-1\\\\\\\frac{5}{12} log_z(5)=-1\qquad \bigg|\;log_a(b)=\frac{log_c(b)}{log_c(a)} \\\\\\\frac{lg(5)}{lg(z)} =-\frac{12}{5} \\\\\\lg(z) = -\frac{5}{12} lg(5)\\\\\\lg(z) = lg(5^{(-5/12)}})\\\\\\z=5^{(-5/12)}\\\\\\z=\frac{1}{5^{(5/12)}}\\\\\\z=5^{(- \displaystyle \frac{5}{12})[/tex]
Ответ:
Объяснение:
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