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Solve (t+2)3/4 =2 where t is a real number.t=

1 Answer

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Solution:

Given the equation:


\begin{gathered} (t+2)^{(3)/(4)}=2 \\ \text{where t is a real number} \end{gathered}

step 1: Take both sides of the equation to the power of 4.

Thus,


\begin{gathered} ((t+2)^{(3)/(4)})^4=(2)^4 \\ \Rightarrow(t+2)^3=16 \end{gathered}

step 2: Take the cube root of both sides of the equation.

Thus,


\begin{gathered} \sqrt[3]{(t+2)^3}=\sqrt[3]{16} \\ \Rightarrow(t+2)=16^{(1)/(3)} \\ \end{gathered}

step 3: Solve for t.


\begin{gathered} 16^{(1)/(3)}^{} \\ \text{can be rewritten as} \\ (2^4)^{(1)/(3)}=2^{(4)/(3)} \\ \text{thus, we have} \\ t+2=2^{(4)/(3)} \\ \text{subtract 2 from both sides of the equation} \\ t+2-2=2^{(4)/(3)}-2 \\ \Rightarrow t=2(2^{(1)/(3)}-1) \\ =2(1.25992-1) \\ \Rightarrow t=2(0.25992) \\ \therefore t=0.51984 \end{gathered}

Hence, the value of t in the equation is evaluated to be 0.51984.

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