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Algebraic Intuition

         

\[ \begin{align*} \large 9^a&\large= 25 \\ \\ \large 3^a &\large = \ ? \end{align*} \]

What goes at the question mark in the equations above?

Hint: There's a much faster way than solving for \(a.\) You are not a robot, and you don't need a calculator to solve this!

\[ \large \frac { 1 }{ { 2 }^{ 3 } } +\frac { 1 }{ { 2 }^{ 6 } } +\frac { 1 }{ { 2 }^{ 9 } } + \cdots = \ ? \]

Hint: Don’t try this one on a calculator either - your fingers will wear out! This algebra problem can be solved with a geometric shortcut:

What total fraction of this image is red?

Which is greater, \(A\) or \(B?\)

\[A=\frac{99^{999}+1}{99^{1000}+1}\]

\[B=\frac{99^{1000}+1}{99^{1001}+1}\]

\(\log_2 3\) \(\times \log_3 4 \) \(\times \log_4 5 \) \(\times \log_5 6 \) \(\times \log_6 7 \) \(\times \log_7 8 = \, ? \)

It’s a race! Who is the winner? Choose wisely...

Which of these sequences will exceed \(1,000,000^{1,000,000}\) first?

Sequence A:\(1^2\)\((1^2)^3\)\(((1^2)^3)^4\)\((((1^2)^3)^4)^5\)...
Sequence B:\(2^1\)\((3^2)^1\)\(((4^3)^2)^1\)\((((5^4)^3)^2)^1\)...
Sequence C:\(2^1\)\(3^{2^1}\)\(4^{3^{2^1}}\)\(5^{4^{3^{2^1}}}\)...

Note: Towers of exponents are evaluated from the top down, so \(3^{2^1} = 3^{\left(2^1\right)}.\)

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