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Which is larger: 4^(6^5) or 5^(3^8)? Question
This is a [calculation-puzzle], which is a puzzle that involves numerical calculations, generally using the basic operations: addition, subtraction, multiplication, and division.
Without using a computer or calculator, determine which number is larger:$$\large4^{6^5} \text{ or }5^{3^8}$$
Rules:
- You cannot use a computer or calculator
- You can use logarithms. To reduce some arithmetic, a limited table of values are supplied (depending on what base log you prefer):$$\begin{array}{|r|c|c|}\hline&\ln x\text{ (base }e\text{)}&\log x\text{ (base }10\text{)}\\\hline2&0.6931&0.3010\\\hline3&1.0986&0.4471\\\hline5&1.6094&0.6989\\\hline\end{array}$$For some additional ease, you can use $\ln(1+x)\approx x-\frac{x^2}2+\frac{x^3}3$ for small $x$ if you need it.
1 answer
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 $$4^{6^5} > 5^{3^8}.$$
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