determining equivalent exponential functions\nwhich exponential functions have been simplified correctly…

determining equivalent exponential functions\nwhich exponential functions have been simplified correctly? choose three correct answers.\n1. $f(x)=81^{\\frac{x}{4}}=3^{x}$\n2. $f(x)=5\\sqrt3{16^{x}}=5(2\\sqrt3{2})^{x}$\n3. $f(x)=\\frac{3}{4}\\sqrt{27^{x}}=\\frac{3}{4}(3\\sqrt{3})^{x}$\n4. $f(x)=(24)^{\\frac{1}{3}x}=2(\\sqrt3{3})^{x}$\n5. $f(x)=2.3(8)^{\\frac{1}{3}x}=2.3(4)^{x}$

determining equivalent exponential functions\nwhich exponential functions have been simplified correctly? choose three correct answers.\n1. $f(x)=81^{\\frac{x}{4}}=3^{x}$\n2. $f(x)=5\\sqrt3{16^{x}}=5(2\\sqrt3{2})^{x}$\n3. $f(x)=\\frac{3}{4}\\sqrt{27^{x}}=\\frac{3}{4}(3\\sqrt{3})^{x}$\n4. $f(x)=(24)^{\\frac{1}{3}x}=2(\\sqrt3{3})^{x}$\n5. $f(x)=2.3(8)^{\\frac{1}{3}x}=2.3(4)^{x}$

Answer

Explanation:

Step1: Verify first function

$81^{\frac{x}{4}} = (3^4)^{\frac{x}{4}} = 3^{4 \cdot \frac{x}{4}} = 3^x$

Step2: Verify second function

$\sqrt[3]{16^x} = (16^x)^{\frac{1}{3}} = (2^4)^x)^{\frac{1}{3}} = 2^{\frac{4x}{3}}$, and $2\sqrt[3]{2} = 2^1 \cdot 2^{\frac{1}{3}} = 2^{\frac{4}{3}}$, so $(2\sqrt[3]{2})^x = 2^{\frac{4x}{3}}$. Thus $5\sqrt[3]{16^x} = 5(2\sqrt[3]{2})^x$

Step3: Verify third function

$\sqrt{27^x} = (27^x)^{\frac{1}{2}} = (3^3)^x)^{\frac{1}{2}} = 3^{\frac{3x}{2}}$, and $3\sqrt{3} = 3^1 \cdot 3^{\frac{1}{2}} = 3^{\frac{3}{2}}$, so $(3\sqrt{3})^x = 3^{\frac{3x}{2}}$. Thus $\frac{3}{4}\sqrt{27^x} = \frac{3}{4}(3\sqrt{3})^x$

Step4: Verify fourth function

$(24)^{\frac{1}{3}x} = (24^{\frac{1}{3}})^x = (\sqrt[3]{24})^x = (\sqrt[3]{8 \cdot 3})^x = 2\sqrt[3]{3}^x$, but $24^{\frac{1}{3}x} \neq 2(\sqrt[3]{3})^x$ (since $\sqrt[3]{24}=2\sqrt[3]{3}$, not $2 \cdot \sqrt[3]{3}$ raised incorrectly)

Step5: Verify fifth function

$8^{\frac{1}{2}x} = (8^{\frac{1}{2}})^x = (\sqrt{8})^x = (2\sqrt{2})^x \neq 4^x$, so $2.3(8)^{\frac{1}{2}x} \neq 2.3(4)^x$

Answer:

  1. $f(x)=81^{\frac{x}{4}}=3^x$
  2. $f(x)=5\sqrt[3]{16^x}=5(2\sqrt[3]{2})^x$
  3. $f(x)=\frac{3}{4}\sqrt{27^x}=\frac{3}{4}(3\sqrt{3})^x$