Do brackets make any difference?

Algebra Level 3

( 5 4 3 ) 2 1 × 5 ( 4 3 ) 2 1 = 5 b ; b = ? \Large\left( \color{#3D99F6}{5}^{\color{#20A900}{4}^{\color{#69047E}{3}}}\right)^{\color{magenta}{2}^{\color{#EC7300}{1}}} \times \color{#3D99F6}{5}^{\left( \color{#20A900}{4}^{\color{#69047E}{3}}\right)^{\color{magenta}{2}^{\color{#EC7300}{1}}}} = \color{#3D99F6}{5}^{\color{#D61F06}{b}} \quad ; \quad \color{#D61F06}{b} = \ ?


The answer is 4224.

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2 solutions

\(\large\begin{array}{} (5^{4^{3}})^{2^{1}} \times 5^{(4^{3})^{2^{1}}} &= (5^{64})^{2} \times 5^{(64^{2})} \\ &= 5^{128} \times 5^{4096} \\ &= 5^{4224} \\ \boxed{b = 4224}\end{array}\)

The answer was nicely symmetric.

Shyambhu Mukherjee - 5 years, 8 months ago

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It's a palindrome.

John Michael Gogola - 5 years, 8 months ago

There is still a lack of parenthesis that make the problem ambiguous (what is 5 4 3 5^{4^3} ?) and I suppose this is why we had multiple answer attempts. ( 5 4 ) 3 5 ( 4 3 ) {(5^4)}^3 \neq 5^{(4^3)}

Anders Therkelsen - 5 years, 8 months ago

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Actually, no ambiguous expression there. It is common to find expressions like these, with several levels of exponentiation. Given that operation is not conmutative, you have to assume (and that's the common agreement for this notation) that 5 4 3 5^{4^3} means 5 ( 4 3 ) 5^{(4^3)} . And, of course, as you point ( 5 4 ) 3 = 5 3 4 = 5 12 5 ( 4 3 ) = 5 2 6 = 5 64 \left(5^4\right)^3=5^{3\cdot 4}=5^{12}\neq 5^{(4^3)}=5^{2^6}=5^{64}

Jose Torres Zapata - 5 years, 8 months ago

Tower rule.

John Michael Gogola - 5 years, 8 months ago

the 2nd one , its 5 ^ 64 ^ 2 ^ 1 after opening the bracket , i dont see the reason for u to solve 64 ^ 2 before 5 ^ 64

Gary Lee - 5 years, 7 months ago
Richard Levine
Jan 7, 2016

Alternatively, (5^64)^2 x 5^(64^2) = 5^64 x 5^64 x 5^(64x64) = 5^(64+64+(64x64)) = 5^(66x64) = 5^b. So b = 66x64 = 4224.

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