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# What is (62^62-62^(62-1))/(62^(62-1))^sqrt ((62^(62-1)+(62^62))-62^(62-2))

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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

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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!

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

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Aug 2, 2017