Is it ok??

Note that $100^{10} = (10^2)^{10} = 10^{20}$ . Now consider $\dfrac {2^{100}}{10^{20}} = \dfrac {2^{100}}{2^{20}5^{20}} = \dfrac {2^{80}}{5^{20}} = \dfrac {16^{20}}{5^{20}} > 1$ , $\implies 2^{100} > 10^{20} = 100^{10}$ . And $\dfrac {3^{70}}{2^{100}} = \dfrac {9^{35}}{2(8^{33})} > \dfrac {9^{35}}{8^{34}} > 1$ , $\implies 3^{70} > 2^{100} > 10^{20} = 100^{10}$ . Therefore, $\boxed{3^{70}}$ is the greatest.

Let 2 $^{100}$ =x

log 2 $^{100}$ =logx

100log2=logx

30.102=logx

10 $^{30.102}$ =x

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Let 3 $^{70}$ =y

log3 $^{70}$ =logy

70log3=logy

33.398=logy

10 $^{33.398}$ =y

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100 $^{10}$ =10 $^{20}$

And definitely 10 $^{20}$ =10 $^{20}$

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Hence, 3 $^{70}$ is greatest

Since we are comparing positive integers, we can take $10^{\text{th}}$ root of all of them. $\sqrt[10]{3^{70}}=3^7=2187 \text{-greatest}$ $\sqrt[10]{2^{100}}=2^{10}=1024$ $\sqrt[10]{100^{10}}=100$ $\sqrt[10]{10^{20}}=10^2=100$ $\therefore \boxed{{3^{70}} \text{is the greatest!}}$

Similar to @Aryan Sanghi 's solution but I want to share it anyway. So we know that one can find the number of digits of a positive integer with a formula

$\#\ digits\ in\ x=\lfloor\lg\left(x\right)\rfloor+1$

We know

$10^{20}=10^{2\cdot10}=\left(10^2\right)^{10}=100^{10}$

And the number of digits:

$\#\ digits\ in\ 10^{20}=\lfloor\lg\left(10^{20}\right)\rfloor+1=\lfloor20\rfloor+1=21$

Let's apply this formula to other numbers too:

$\#\ digits\ in\ 2^{70}=\lfloor\lg\left(2^{70}\right)\rfloor+1=31$

$\#\ digits\ in\ 3^{70}=\lfloor\lg\left(3^{70}\right)\rfloor+1=34$

So from this, we can conclude that $3^{70}$ is the largest