Exponent Towers

Algebra Level 3

Find $x$ , satisfying

$\Large 2^{2^{3^{2^{2}}}} = 4^{4^{x}}.$

2 solutions

André Hucek
Oct 13, 2017

RHS can be simplified as $4^{4^{x}} = (2^2)^{(2^{2)^{x}}} = 2^{2 \times 2 ^{2x}} = 2^{2^{\color{#D61F06}{2x+1}}}$ .

So if $2x+1 = 3^{2^{2}} = 3^4 = 81$

$x = \boxed{40}$

We get, 2^24=2^8x. So, x = 3. Is it wrong?

Anurup Sinha - 3 years, 7 months ago

$x$ is in your RHS, start there and simplify so that it satisfies LHS

André Hucek - 3 years, 7 months ago

Calvin Lin Staff - 3 years, 7 months ago

T.J. Span
Nov 13, 2017

The problem is fun to solve using logarithm rules too.

$2^{2^{3^{2^{2}}}}=4^{4^{x}}$

$log_{2}2^{2^{3^{2^{2}}}}=log_{2}4^{4^{x}}$

$2^{3^{2^{2}}}log_{2}2=4^{x}log_{2}4$

$2^{3^{2^{2}}}=2\cdot 4^{x}$

$log_{2}2^{3^{2^{2}}}=log_{2}(2\cdot 4^{x})$

$3^{2^{2}}log_{2}2=log_{2}2+x\cdot log_{2} 4$

$3^{2^{2}}=1+2x$

$2x=3^{2^{2}}-1$

$2x=80$

${\color{teal} x=40}$