Which of these numbers is the biggest

Question

How do I determine algebraically/without a calculator which of these numbers is the greatest?

$$8^{36}, 7^{55}, 5^{72}, 2^{110}$$

Please provide the method and a little description for me to understand the method properly.

Answer 1

Here is a more direct, comparison-based heuristic:

1. $8^{36}$ vs. $2^{110}$: Since $8=2^3$, we get $8^{36}=(2^3)^{36}=2^{108}<2^{110}$.
2. $2^{110}$ vs. $5^{72}$: Since $2^2=4<5$, we get $2^{110}=4^{55}<5^{72}$.
3. $5^{72}$ vs. $7^{55}$: I'm afraid I'll deviate a bit from the prior two steps here, but consider the fraction $5^{72}/7^{55}=(5/7)^{55}\cdot 5^{17}$. The trick I want to use now is using the 17 remaining 5's to make all factors greater than 1, thus convincing me that $5^{72}>7^{55}$. To do this, I compute the largest power $n$ such that $(5/7)^n>1/5$, so that $5(5/7)^n>1$. This I did by brute-force, no calculator. I got that $n=4$ yields $(5/7)^4=625/2401$ and $5(5/7)^4=3125/2401>1$. So do I have enough 5's to distribute to the 55 fractions? We need one 5 for each four (5/7) factors. Luckily, $55/4=13.75<17$, there are enough 5's, and we can conclude.

Answer 2

To get the number of digits you can use the following formula:

$$\text{Number of digits of}\space\space x=\lfloor\log_{10}\left(x\right)\rfloor+1$$

Notice also that, assuming $a,b\in\mathbb{R}^+$:

$$\log_{10}\left(a^b\right)=\frac{b\ln(a)}{\ln(10)}\approx\frac{b\ln(a)}{2.30258509299}$$

---

So:

• $8^{36}$ has this number of digits: $$\lfloor\log_{10}\left(8^{36}\right)\rfloor+1=\lfloor\frac{36\ln(8)}{\ln(10)}\rfloor+1=33$$
• $7^{55}$ has this number of digits: $$\lfloor\log_{10}\left(7^{55}\right)\rfloor+1=\lfloor\frac{55\ln(7)}{\ln(10)}\rfloor+1=47$$
• $5^{72}$ has this number of digits: $$\lfloor\log_{10}\left(5^{72}\right)\rfloor+1=\lfloor\frac{72\ln(5)}{\ln(10)}\rfloor+1=51$$
• $2^{110}$ has this number of digits: $$\lfloor\log_{10}\left(2^{110}\right)\rfloor+1=\lfloor\frac{110\ln(2)}{\ln(10)}\rfloor+1=34$$

So the biggest number in your list is: $5^{72}$

Answer 3

\begin{align*} & \mathrel{\phantom{<}} 8^{36} = (2^3)^{36} = 2^{108} \\ & < 2^{110} =(2^2)^{55} = 4^{55} \\ & < 7^{55} < 7^{56} = (7^4)^{14} = 2401^{14} < 3125^{14} = (5^5)^{14} = 5^{70} \\ & < 5^{72}. \end{align*}

Answer 4

There are two things varying here: The base and the exponent. For instance in $5^{72}$, $5$ is the base and $72$ is the exponent. The larger either of them are, the larger the number.

Here are the numbers in base $10$:

\begin{align} 8^{36} \approx 3.2 \cdot 10^{32} \\ 7^{55} \approx 3.0 \cdot 10^{46} \\ 5^{72} \approx 2.1 \cdot 10^{50} \\ 2^{110} \approx 1.3 \cdot 10^{33} \end{align}

so $5^{72}$ is the biggest one.