$a \log_bx=\log_{b^\frac{1}{a}}x$
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I think this is my fault. I meant to write the base alone is raised to the power of 1/a, not log_b in the second equation.
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Yes! I revised my solution. The statement is true. Let me know if you need any clarifications.
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Thank you Erdos that explains it perfectly
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I edited the title accordingly.
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