Nilai x dari persamaan [tex]{\sf{3^{3x+\frac{1}{2}}=\sqrt{\left(\dfrac{\sf{1}}{\sf{243}}\right)^{\sf{x-4}}}}}[/tex] adalah [tex]{\boldsymbol{\dfrac{\textsf{\textbf{19}}}{\textsf{\textbf{11}}}}}[/tex]
Pembahasan
[tex]\begin{aligned}{\sf{3^{3x+\frac{1}{2}}}}&={\sf{\sqrt{\left(\frac{1}{243}\right)^{x-4}}}}\\{\sf{3^{3x+\frac{1}{2}}}}&={\sf{\sqrt{\left(\frac{1}{3^5}\right)^{x-4}}}}\\{\sf{3^{3x+\frac{1}{2}}}}&={\sf{\sqrt{\left(3^{-5}\right)^{x-4}}}}\\{\sf{3^{3x+\frac{1}{2}}}}&={\sf{\sqrt{3^{-5x+20}}}}\\{\sf{/\!\!\!3^{3x+\frac{1}{2}}}}&={\sf{/\!\!\!3^{\frac{-5x+20}{2}}}}\\{\sf{3x+\frac{1}{2}}}&={\sf{\frac{-5x+20}{2}}}\\{\sf{2\left(3x+\frac{1}{2}\right)}}&={\sf{2\left(\frac{-5x+20}{2}\right)}}\\{\sf{6x+1}}&={\sf{-5x+20}}\\{\sf{6x+5x}}&={\sf{20-1}}\\{\sf{11x}}&={\sf{19}}\\{\sf{x}}&={\sf{\frac{19}{11}}}\end{aligned}[/tex]
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