It's from an exam problem mostly, however, I believe will help many others out here.
Problem #1
Which one is bigger: $31^{11}$ OR $17^{14}$
Problem #2
Which one is bigger: $31^{11}$ OR $14^{14}$
My logarithmic way for first one: $31^{11}$ ? $17^{14} \rightarrow 31 ? \;17^{14/11} \rightarrow 31 ?\; (17\cdot17^{0.3}$).
So $31^{11}$ < $17^{14}$. However, the problem with this way is the $17^{0.3}$, which I can't calculate without a calculator.
So Problem #3
How $17^{0.3}$ can be calculated without a calculator (while assuming I've memorized the values of $\log 2,\log 3,\log 5$ and $\log 7$.)
Please mention if there's any general way to solve these problems, fast!
Thank you!
Best Answer
\begin{align*} \frac{31^{11}}{17^{14}}&=\frac{31^{11}}{17^{11}.17^{3}}\\ &=\left(\frac{31}{17} \right)^{11}\frac{1}{17^3}\\ &<\frac{2^{11}}{17^3}\\ &<\frac{2^{11}}{16^3}\\ &=\frac{2^{11}}{2^{12}}\\ &=\frac{1}{2}\\ &<1 \end{align*} So $\frac{31^{11}}{17^{14}}<1$. Multiply both sides by $17^{14}$ and you get $31^{11}<17^{14}$