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Mathematics

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navik [9.2K]2 years ago

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Answer:

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vichka [17]

Answer:

$\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}$

Step-by-step explanation:

$(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}$

$2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------$

$3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}$