Log problem

Algebra Level 4

log 5 x = x [ 5 1 ( 1 + x 5 1 ) ] 5 5 1 , x = ? \Large\log_5x=x^{\left[5^{-1}(1+x^{-5^{-1}})\right]}5^{-5^{-1}}\qquad,\qquad x = \ ?

Solve for positive integer x x .


The answer is 3125.

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1 solution

Ben Habeahan
Sep 3, 2015

Let x = p 5 x=p^5 or p = x 1 5 . p=x^{ \frac{1}{5}}.

After substitute we have, log 5 p 5 = p 5 [ 1 5 ( 1 + p 1 ) ] 5 1 5 . \log_{5}{p^5}={p^5}^{ [\ \frac{1}{5}(1+p^{-1})]\ }5^ \frac{-1}{5}.

( ( 5 ) ( 5 1 5 ) ) log 5 p = p ( 1 + p 1 ) \Leftrightarrow {((5)(5^ \frac{1}{5}))} \log_{5}{p}={p}^{(1+ p^{-1})}

log 5 p 5 ( 6 5 ) = p ( 1 + 1 p ) \Leftrightarrow \log_{5}{p^{5^{ (\frac{6}{5})}}}={p}^{(1+\frac{1}{p})}

p 6 = 5 p ( 1 + 1 p ) \Leftrightarrow p^{6}= 5^{p^{(1+\frac{1}{p})}}

p 6 = 5 ( p + 1 ) \Leftrightarrow p^{6}= 5^{(p+1)}

With o b s e r v a t i o n , \color{#3D99F6}{observation}, the statement p 5 + 1 = 5 ( p + 1 ) p^{5+1}= 5^{(p+1)} true iff p = 5. p=5.

x = p 5 = 5 5 = 3125 x=p^5=5^5= \boxed{3125}

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