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if 9^x-9^x-1=24, then find the value of (6x)^x
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That equation cannot possibly be true because : \[9^x-9^x=0\] and 0-1=24 can never be true for any value of x.
actually it CAN!! not really.
Well, maybe if you used the premise that the reflexive axiom was invalid.
First find \(x\): \[\begin{array}{rcl} 9^x-9^{x-1}&=&24\\ 9^x-\frac{9^x}{9}&=&24 \\ \frac{8}{9}9^x&=&24 \\ 9^x &=& \frac{9\cdot 24}{8}\\ 9^x &=& 9\cdot \sqrt{9} \\ x&=& \frac{3}{2} \end{array} \] Then plug it into \((6x)^x\)\[ \begin{split} (6x)^x &= \left( \frac{6 \cdot 3}{2} \right)^{3/2} \\ &=9^{3/2}\\ &= 27 \end{split} \]
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