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Suppose $81^x = 64$. what is $27^{x 1}$?
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Suppose $81^x = 64$. what is $27^{x 1}$?

User Aram Becker
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1 Answer

7 votes
x ln 81 = ln 64

x = ln 64/ln 81
x+1 = (ln 64 +ln 81)/ln 81

27^((ln 64+ln 81)/ln 81)
and simplify hope this helps!
User Jfmatt
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