Welcome to the staging ground for new communities! Each proposal has a description in the "Descriptions" category and a body of questions and answers in "Incubator Q&A". You can ask questions (and get answers, we hope!) right away, and start new proposals.
Are you here to participate in a specific proposal? Click on the proposal tag (with the dark outline) to see only posts about that proposal and not all of the others that are in progress. Tags are at the bottom of each post.
Post History
Let's first look at the exponents: \begin{align} 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\ 3^8 &= 3^3\cdot 3^5 \end{align} We see that both exponents have a factor of $3^5$, which we can th...
#4: Post edited
by
celtschk
·
2025-03-19T18:06:39Z (about 1 month ago)
Copied clarifying formulas from comment to answer body
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity of the power functions. The question therefore reduces to: What is larger, $4^{2^5}=(2^2)^{32}=2^{64}$ or $5^{3^3}=5^{27}$?- Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^{27}$ to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
- We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity of the power functions:
- \begin{align}
- a^{bn}&<c^{dn}\\\\
- \implies (a^b)^n&<(c^d)^n && \text{(power law)}\\\\
- \implies a^b &< c^d && \text{(monotonicity of power function)}
- \end{align}
- The question therefore reduces to: What is larger, $4^{2^5}=(2^2)^{32}=2^{64}$ or $5^{3^3}=5^{27}$?
- Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^{27}$ to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
#3: Post edited
by
celtschk
·
2025-03-19T18:04:21Z (about 1 month ago)
Wording and spelling error
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity in the exponent. The quesion therefore reduces to: What is larger, $4^{2^5}=(2^2)^{32}=2^{64}$ or $5^{3^3}=5^{27}$?- Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^{27}$ to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
- We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity of the power functions. The question therefore reduces to: What is larger, $4^{2^5}=(2^2)^{32}=2^{64}$ or $5^{3^3}=5^{27}$?
- Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^{27}$ to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
#2: Post edited
by
celtschk
·
2025-03-16T20:56:00Z (about 1 month ago)
Fixed a type and streamlined the argument
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity in the exponent. The quesion therefore reduces to: What is larger, $4^{2^5}=4^{32}$ or $5^{3^3}=5^{27}$?Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^27$ and make use of $4=2^2$, that is, $4^32 =2^{64}$, to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
- Let's first look at the exponents:
- \begin{align}
- 6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
- 3^8 &= 3^3\cdot 3^5
- \end{align}
- We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity in the exponent. The quesion therefore reduces to: What is larger, $4^{2^5}=(2^2)^{32}=2^{64}$ or $5^{3^3}=5^{27}$?
- Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^{27}$ to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
- Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
- $$4^{6^5} > 5^{3^8}.$$
#1: Initial revision
by
celtschk
·
2025-03-16T20:53:31Z (about 1 month ago)
Let's first look at the exponents:
\begin{align}
6^5 = (2\cdot 3)^5 &= 2^5\cdot 3^5\\\\
3^8 &= 3^3\cdot 3^5
\end{align}
We see that both exponents have a factor of $3^5$, which we can therefore drop because of the monotonicity in the exponent. The quesion therefore reduces to: What is larger, $4^{2^5}=4^{32}$ or $5^{3^3}=5^{27}$?
Noting that we have $5$ to the power of a multiple of $3$, and compare it with a power of $2$, maybe we can make use of the well-known fact that $10^3=1000$ is slightly smaller than $2^{10}=1024$. Therefore, let's multiply both sides with $2^27$ and make use of $4=2^2$, that is, $4^32 =2^{64}$, to see that we have to compare the numbers $2^{64+27} = 2^{91}$ and $10^{27} < 2^{90}$.
Clearly $2^{91}$ is larger than $2^{90}$ and therefore than $10^{27}$. Therefore we can conclude that
$$4^{6^5} > 5^{3^8}.$$