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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!

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    $\begingroup$ Do you mean "or"? or else the problem doesn't make sense $\endgroup$ – suomynonA Nov 3 '16 at 4:48
  • $\begingroup$ Yes, corrected! $\endgroup$ – neo-nant Nov 3 '16 at 5:16
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For the second problem, We might notice that $\frac{31}{14} > 2$ and $2^4 = 16 > 14$, suggesting the following reasoning:

$$ 31^{11} > 28^{11} = 2^{11}\cdot14^{11} = 2^3 \cdot 16^2\cdot14^{11} > 8 \cdot 14^{13}. $$ The rightmost quantity falls short of $14^{14}$ by a factor of nearly $2$. But we can scrounge up another factor of $2$ from the left-hand end. Observing that $31 > 1.1 \cdot 28$, we have

$$ 31^{11} > 1.1^{11}\cdot 28^{11} > 2 \cdot (2^{11}\cdot 14^{11}) = 2^{12}\cdot14^{11} = 16^3\cdot14^{11} > 14^{14}. $$

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  • $\begingroup$ All the answers are great.Yet I can't help but choose yours! $\endgroup$ – neo-nant Nov 3 '16 at 6:37
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\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 $31^{11}<17^{14}$

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  • $\begingroup$ Thanks! Could you please shoot a few lines for no 2? $\endgroup$ – neo-nant Nov 3 '16 at 5:58
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    $\begingroup$ I prefer this answer to the chosen one, simply because this one describes an algorithmic way of solving almost any other similar problem: start by writing the ratio and then keep getting the two numbers closer and closer through comparison with powers that are easier to work with (e.g., transforming 17 into 16). $\endgroup$ – Bogdan Alexandru Nov 3 '16 at 8:47
  • $\begingroup$ @BogdanAlexandru, I really like this way of solving the 1st problem. But I could not solve the 2nd problem with this way rather via accepted one, more easily. Regards. $\endgroup$ – neo-nant Nov 3 '16 at 16:33
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$31^{11} < 32^{11} = 2^{11} 16^{11} < 2^{12} 17^{11} = 16^{3} 17^{11} < 16^3 17 ^{11} < 17^{14}$.

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  • $\begingroup$ Thanks! I would love your second's answer. $\endgroup$ – neo-nant Nov 3 '16 at 6:39
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    $\begingroup$ This could be simplified a little bit: $31^{11} < 32^{11} = 2^{55} < 2^{56} = 16^{14} < 17^{14}$ $\endgroup$ – Philip C Nov 3 '16 at 9:12
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Given your reference values, you can use

$$ \log(16) < \log(17) < \log(18) $$ $$ \log(30) < \log(31) < \log(32) $$

to get upper and lower bounds on the logarithm.

The first you could actually skip computing the values altogether:

$$ 11 \log(31) < 11 \log(32) = 55 \log(2) < 56 \log(2) = 14 \log(16) < 14 \log(17)$$

but for the other problems, using the upper and lower bounds on $\log(17)$ and $\log(31)$ give enough precision to solve the problem.

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Problems #1 and #2 were already addressed by other answers; I would like to give my insight for Problem #3:

How can $a^b$ be calculated without a calculator, assuming $0 < b < 1$ and assuming I've memorized that $$\log 2 \approx 0.7$$ $$\log 3 \approx 1.1 $$ $$\log 5 \approx 1.6$$ $$\log 7 \approx 1.95$$

I will consider the example $a = 17$ and $b = 0.3$.

Step 1. Calculate an approximation to $\log a$, using the logarithms you memorized:

How to calculate an approximation to $\log 17$ ? Find a way to express $17$ as a product (or fraction) of powers of the ones that you memorized. See some examples below, choose the one that you like most:

$$\log 17 \approx \log (2^4)$$ $$\log 17 \approx \log (2 \cdot 3^2)$$ $$\log 17 \approx \log \dfrac{5 \cdot 7}{2}$$ $$\log 17 \approx \log \dfrac{2 \cdot 5^2}{3}$$ $$\log 17 \approx \log \dfrac{2^2 \cdot 5^4}{3 \cdot 7^2}$$

Taking the last one (the best among those), we get

$$\log 17 \approx \log \dfrac{2^2 \cdot 5^4}{3 \cdot 7^2} \approx 2 \cdot 0.7 + 4 \cdot 1.6 - 1.1 - 2 \cdot 1.95 = 2.8$$

Step 2. Calculate $a^b$ using it's MacLaurin expansion (with the approximation from step 1) until you get tired:

Let $x = b \log a$. In our example, $x = 0.3 \cdot 2.8 = 0.84$.

Now use the MacLaurin expasion, truncated to as many terms as you like:

$$a^b = e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} + \dfrac{x^6}{6!} + \dfrac{x^7}{7!} + \cdots$$

In our example, truncated to the fourth term, we have

$$17^{0.3} = 1 + 0.84 + \dfrac{0.84^2}{2} + \dfrac{0.84^3}{3!} = 2.292$$

While the correct value is $17^{0.3} = 2.33956263$. I would say it's good enough...

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Since the log function is monotonically increasing, the following is true: For positive $x$, $y$, $\log x > \log y$ implies $x > y$. It doesn't matter which base we use, but note that 17 and 31 are both nearly powers of 2, so use base 2.

We'll work with the ratio of the logs to determine if the ratio is greater than 1 (meaning the numerator is greater) or less than one (meaning the denominator is greater).

$$ \begin{array}{tcl} \frac{\lg 31^{11}}{\lg 17^{14}} & = & \frac{11\lg 31}{14\lg 17} \\ & < & \frac{11\lg32}{14\lg16} \\ & = & \frac{11(5)}{14(4)} \\ & = & \frac{55}{56} \\ & < & 1 \end{array} $$

We can justify the first inequality by noting that the ratio increases if we either increase the numerator or decrease the denominator, and we've done both.

Since the ratio is less than one, the denominator is bigger, meaning $17^{14} > 31^{11}$.

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