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What's (62^62-62^(62-1))/(62^(62-1))^sqrt ((62^(62-1)+(62^62))-62^(62-2))

 Aug 2, 2017
 #1
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Wow! That is such a mouthful of an expression, but I will attempt to evaluate it anyway. 

 

If I am not mistaken, the expression is the following:
 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\)

 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\) I'll start with the fraction portion of this expression
\(\frac{62^{62}-62^{62-1}}{62^{62-1}}\) First, I will break the fraction up with the following fraction rule that states \(\frac{a\pm b}{c}=\frac{a}{c}\pm\frac{b}{c}\).
\(\frac{62^{62}-62^{62-1}}{62^{62-1}}=\frac{62^{62}}{62^{62-1}}-\frac{62^{62-1}}{62^{62-1}}\) Of course, \(\frac{62^{62-1}}{62^{62-1}}=1 \) because any nonzero number divided by itself is always one.
\(\frac{62^{62}}{62^{62-1}}-\frac{62^{62-1}}{62^{62-1}}=\frac{{62^{62}}}{62^{61}}-1\) We will utilize an exponent rule that states that \(\frac{a^b}{a^c}=a^{b-c}\).
\(\frac{{62^{62}}}{62^{61}}=62^{62-61}=62^1=62\) Reinsert this in for \(\frac{{62^{62}}}{62^{61}}\).
\(62-1=61\) Ok, we have successfully simplified from \(\frac{62^{62}-62^{62-1}}{62^{62-1}}\) to 61. Reinsert 61 for \(\frac{62^{62}-62^{62-1}}{62^{62-1}}\) in the original expression.
\(61^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\) I will convert \(62^{62-1}\) into a fraction by using the converse of a fraction rule I utilized before. It is \(a^{b-c}=\frac{a^b}{a^c}\)
\({\sqrt{62^{62-1}+62^{62}-62^{62-2}}}={\sqrt{\frac{62^{62}}{62^1}+62^{62}-62^{62-2}}}\) Multiply \(62^{62}\) by \(\frac{62}{62}\) to create a common denominator.
\(\frac{62^{62}}{1}*\frac{62}{62}=\frac{62*62^{62}}{62^1}\) Insert this back into the equation, too.
\({\sqrt{\frac{62^{62}}{62^1}+62^{62}-62^{62-2}}}={\sqrt{\frac{62^{62}}{62^1}+\frac{62*62^{62}}{62^1}-62^{62-2}}}\) Add \(\frac{62^{62}}{62^1}+\frac{62*62^{62}}{62^1}\) together. 
\({\sqrt{\frac{62^{62}}{62^1}+\frac{62*62^{62}}{62^1}-62^{62-2}}}=\sqrt{\frac{63*62^{62}}{62^1}-62^{60}}\) Convert \(\frac{62^{60}}{1}*\frac{62}{62}\) into a fraction over 62 by doing
\(\frac{62^{60}}{1}*\frac{62}{62}=\frac{62^{60}*62}{62}=\frac{62^{61}}{62}\) Replace   \(62^{60}\) with \(\frac{62^{61}}{62}\)
\(\sqrt{\frac{63*62^{62}}{62^1}-62^{60}}=\sqrt{\frac{63*62^{62}}{62^1}-\frac{62^{61}}{62}}\) Subtract the fractions.
\(\sqrt{\frac{63*62^{62}}{62^1}-\frac{62^{61}}{62}}=\sqrt{\frac{63*62^{62}-62^{61}}{62}}\) "Distribute" the square root to both the numerator and denominator.
\(\sqrt{\frac{63*62^{62}-62^{61}}{62}}=\frac{\sqrt{63*62^{62}-62^{61}}}{\sqrt{62}}\) Rationalize the denominator by multiplying \(\frac{\sqrt{63*62^{62}-62^{61}}}{\sqrt{62}}\) by \(\frac{\sqrt{62}}{\sqrt{62}}\).
\(\frac{\sqrt{63*62^{62}-62^{61}}}{\sqrt{62}}*\frac{\sqrt{62}}{\sqrt{62}}\) Use the square root rule that states that \(\sqrt{x}*\sqrt{y}=\sqrt{xy}\)
\(\frac{\sqrt{62(63*62^{62}-62^{61})}}{62}\) Distribute the 62 to all terms. 
\(\frac{\sqrt{62(63*62^{62}-62^{61})}}{62}=\frac{\sqrt{62*63*62^{62}-62*62^{61}}}{62}\) Simplify.
\(\frac{\sqrt{3906*62^{62}-62^{62}}}{62}\) Do the subtraction in the numerator.
\(3906*62^{62}-62^{62}=3905*62^{62}\) Reinsert that back into the expression. I will simplify \(\sqrt{3905*62^{62}}\) by using the rule that \(\sqrt{ab}=\sqrt{a}*\sqrt{b}\)
\(\sqrt{3905}*\sqrt{62^{62}}\)  
\(\sqrt{62^{62}}=\sqrt{62^{31}*62^{31}}=62^{31} \) Plug this back into the original expression
\(\frac{62^{31}\sqrt{3905}}{62}=62^{30}*\sqrt{3905}\) Factor out the common factor of 62
   


At this point, I cannot simplify the exponent any further. I have done an impressive job of simplifying 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}\) into \(61^{({62^{30}*\sqrt{3905}})}\). I will now have to utilize an extremely accurate calculator for a calculation of this size. My only option is to provide you with an exponent tower; that's how huge it is!

 

\((\frac{62^{62}-62^{62-1}}{62^{62-1}})^{\sqrt{62^{62-1}+62^{62}-62^{62-2}}}=61^{({62^{30}\sqrt{3905}})}\approx 10^{10^{55.81927966668297}}\)

.
 Aug 2, 2017

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