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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...
Answer
#4: Post edited
by
celtschk
·
2025-03-19T18:06:39Z (12 days 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 (12 days 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 (15 days 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 (15 days 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
$$4^{6^5} > 5^{3^8}.$$