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30 Nov 2021, 04:06
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If \(\sqrt{\sqrt{\sqrt{5x}}} = \sqrt[6]{4x}\), what is the range of the possible values of \(x\)?
A. \(0\)
B. \(\frac{1}{4}\)
C. \(\frac{125}{256}\)
D. \(\frac{1}{2}\)
E. \(\frac{131}{256}\)
A. \(0\)
B. \(\frac{1}{4}\)
C. \(\frac{125}{256}\)
D. \(\frac{1}{2}\)
E. \(\frac{131}{256}\)
30 Nov 2021, 04:06
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Official Solution:
If \(\sqrt{\sqrt{\sqrt{5x}}} = \sqrt[6]{4x}\), what is the range of the possible values of \(x\)?
A. \(0\)
B. \(\frac{1}{4}\)
C. \(\frac{125}{256}\)
D. \(\frac{1}{2}\)
E. \(\frac{131}{256}\)
\(\sqrt{\sqrt{\sqrt{5x}}} = \sqrt[6]{4x}\);
\((5x)^{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} =(4x)^{\frac{1}{6}}\);
\((5x)^{\frac{1}{8}} =(4x)^{\frac{1}{6}}\);
Take to the power of 24 (the LCM of 8 and 6): \((5x)^3 =(4x)^4\);
\(x^3(256x-125)=0\);
\(x=0\) or \(x=\frac{125}{256}\).
The range of the possible values of \(x\) is therefore \(\frac{125}{256}-0=\frac{125}{256}\).
Answer: C
If \(\sqrt{\sqrt{\sqrt{5x}}} = \sqrt[6]{4x}\), what is the range of the possible values of \(x\)?
A. \(0\)
B. \(\frac{1}{4}\)
C. \(\frac{125}{256}\)
D. \(\frac{1}{2}\)
E. \(\frac{131}{256}\)
\(\sqrt{\sqrt{\sqrt{5x}}} = \sqrt[6]{4x}\);
\((5x)^{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} =(4x)^{\frac{1}{6}}\);
\((5x)^{\frac{1}{8}} =(4x)^{\frac{1}{6}}\);
Take to the power of 24 (the LCM of 8 and 6): \((5x)^3 =(4x)^4\);
\(x^3(256x-125)=0\);
\(x=0\) or \(x=\frac{125}{256}\).
The range of the possible values of \(x\) is therefore \(\frac{125}{256}-0=\frac{125}{256}\).
Answer: C
27 Apr 2024, 09:39
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Thanks a lot for answering this question Bunuel, I am not able to understand aspect of "Take to the power of 24" can you please help understand that concept
Thank you
Thank you
