Indices (Equation)

 

CIE  \0606 \2020 \w \ 22 \ No 2

Find the value of $x$ such that $\dfrac{4^{x+1}}{2^{x-1}}=32^{\dfrac{x}{3}} \times 8^{\dfrac{1}{3}}$.


*********math solution************* ****
$\begin{array}[t]{rcll} \dfrac{\left(2^{2}\right)^{x+1}}{2^{x-1}} &=&\left(2^{5}\right)^{\dfrac{x}{3}} \times\left(2^{3}\right)^{\dfrac{1}{3}} \\2^{2(x+1)-(x-1)}&=& 2^{\dfrac{5 x}{3}+1}\\ 2 x+2-x+1&=& \dfrac{5 x}{3}+1 \\3(x+3)&=&5 x+3 \\3 x+9&=&5x+3 \\2 x&=&6 \\x&=&3\end{array}$ ****


**********end math solution********************$

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