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I'm required to compare $e^{4}-2$ and $50$ without using calculator. I thought of the following way:

Let a function $h(x)$ be defined as $e^{x}-13x.$ If I can prove that this function at $x=4$ is positive(without the use of calculator again) then the job is done. But I'm not able to do so. How to proceed?

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    $\begingroup$ $e^4 = 1+\frac{4}{1}+\frac{4^2}{2}+\frac{4^3}{6}+\frac{4^4}{24}+\frac{\xi^5}{120}$ etc to as many terms you need to reach 50 $\endgroup$
    – fGDu94
    May 23 '21 at 17:45
  • $\begingroup$ $h(4)=e^4-50,$ and you want to ask about $e^4-52.$ $\endgroup$ May 23 '21 at 17:49
  • $\begingroup$ Even to $4^7/7!$ you don’t get a good estimate of $e^4.$ @fGDu94 $\endgroup$ May 23 '21 at 17:54
  • $\begingroup$ @ThomasAndrews $2.7<e$ and $(2.7)^4-2>50$ $\endgroup$
    – Logic
    May 23 '21 at 17:56
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    $\begingroup$ $e^4 > 2.7^4 = (2+0.7)^4 = (4+2.8+0.49)^2 = (7.29)^2 = 49+7*0.58+0.29^2 > 50$ might be more suitable $\endgroup$
    – fGDu94
    May 23 '21 at 17:56
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Use:

$$e^2>1+\frac21+\frac{2^2}2+\frac{2^3}6+\frac{2^4}{24}+\frac{2^5}{120}=7 +\frac{4}{15}.$$ So $$e^4>\left(7+\frac4{15}\right)^2>7^2+2\cdot 7\cdot \frac{4}{15}>52.$$

The last step because: $$2\cdot 7\cdot\frac4{15} =(15-1)\cdot \frac4{15}=4-\frac{4}{15}>3.$$

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Note that $(2.7)^4$ is greater than $52$ and since $e>2.7$ it’s fourth power must also be greater than $52$.

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Alternative approach: use logarithms, base $10$.

Warning: this is long-winded.

The challenge is to do everything without a calculator.

For that, you will need to derive the base $(10)$ logarithms of $(2), (3), (7), (11), (13),$ and $(e = 2.718+)$, without a calculator.


The approach taken will be:

  • Step 1: use a somewhat bizarre form of numerical interpolation to derive a rough estimate for $\log_{10} (e)$.

  • Step 2: use this estimate to compute the logarithms base $(10)$ for $2,3,11$, each accurate to 3 decimal places.

  • Step 3: use these logarithms to refine the $\log_{10} (e)$.

  • Step 4: derive $\log_{10} (7)$ and then $\log_{10} (13).$

  • Step 5: compare $\log_{10} (e^4)$ with $\log_{10} (4 \times 13).$


$\underline{\text{Step 1:}}$

Let $f(x) = \log_{10}(x)~~$ and $~~g(x) = \ln(x) = \log_{e}(x).$

Suppose that $~a = \log_{10}(x), ~b = \log_{e}(x), ~c = \log_{10}(e).$
Then $(10)^{a} = x = e^b = \left[(10)^c\right]^b = (10)^{(bc)}.$
Therefore, $a = bc$.
Thus, $f(x) = [\log_{10}{(e)}] \times g(x)$.
Therefore, $f'(x) = [\log_{10}{(e)}] \times g'(x) = [\log_{10}{(e)}] \times \dfrac{1}{x}.$

$2^{10} \approx 10^3 \implies \log_{10}(2) \approx (0.3) \implies \log_{10}(4) \approx (0.6).$
Therefore, as $x$ goes from $(2)$ to $(4), ~f(x)$ changes from $\approx (0.3)$ to $\approx (0.6)$.

The (rough) average rate of change between $2$ and $3$ for $f(x)$ will be
$\approx [\log_{10}{(e)}] \times \dfrac{(1/2) + (1/3)}{2}.$

Similarly, the (rough) average rate of change between $3$ and $4$ for $f(x)$ will be
$\approx [\log_{10}{(e)}] \times \dfrac{(1/3) + (1/4)}{2}.$

This implies that $f(4) - f(2)$ may be estimated as

$\approx [\log_{10}{(e)}] \times \dfrac{(2/3) + (3/4)}{2} \approx [\log_{10}{(e)}] \times \dfrac{7}{10}.$

This implies that an initial rough estimate for $[\log_{10}{(e)}]$ is $\dfrac{(3/10)}{(7/10)} = \dfrac{3}{7}.$


$\underline{\text{Step 2:}}$

$2^{(10)} = 1024.$
Therefore, to refine the computation of $\log_{10}(2)$, you need a reasonably accurate estimate of $f(1024) - f(1000).$

The average rate of change, for $f(x)$ in the interval $[1000, 1024]$ will be $\approx \dfrac{3}{7} \times \dfrac{1}{1000}.$

Since the interval $[1000, 1024]$ is $(24)$ units wide, you have that
$f(1024) - f(1000) \approx \dfrac{3}{7} \times \dfrac{1}{1000} \times (24) \approx \dfrac{72}{7} \times \dfrac{1}{1000} \approx \dfrac{1}{100}.$

Therefore $\log_{10}(1024) \approx 3 + \dfrac{1}{100}.$

Therefore, since $2^{(10)} = 1024, ~~\log_{10}(2) \approx (0.301).$

.....

Similarly, you know that $3^4 = 81$ and that $\log_{10}(80) \approx (1.903).$

$f(81) - f(80)$ may be estimated as
$\dfrac{3}{7} \times \dfrac{1}{80} = \dfrac{3}{560} \approx \dfrac{5}{1000}.$

This implies that $\log_{10}(81) \approx (1.903) + (0.005) = (1.908)$.
Therefore $\log_{10}(3) \approx (0.477)$.

.....

From Pascal's triangle, you know that $(11)^3 = (1331)$
which is very close to $1333 + \dfrac{1}{3} = (4/3) \times (1000).$

You now know that $\log_{10}[(4/3) \times (1000)] \approx (3.602 - 0.477) = (3.125).$

Further, $f[(4/3) \times (1000)] - f(1331) \approx \dfrac{7}{3} \times \dfrac{3}{7} \times \dfrac{1}{(4/3) \times 1000} \approx 0.00075.$

Therefore, $\log_{10}(1331) \approx (3.125) - (0.00075) = (3.12425)$.

Therefore $\log_{10}(11) \approx (1.0414)$.


$\underline{\text{Step 3:}}$

$e = 2.71828+$ which is approximately $(2/3)$ of the way from
$\left(2.70 = \dfrac{3^3}{10}\right)$ to $\left(2.7\overline{27} = \dfrac{30}{11}\right).$

From Step 2, you have that

  • $\log_{10} \dfrac{3^3}{10} \approx (0.431)$.

  • $\log_{10} \dfrac{30}{11} \approx (1.4770 - 1.0414) \approx (0.4356).$

The interval between $(0.431)$ and $(0.4356)$ is about $(0.0046).$
$(2/3)$ of this interval is about $(0.003)$.

Taking the weighted average, (2/3) of the way along this interval, you therefore have that $\log_{10}(e) \approx (0.431) + (0.003) = (0.434).$

Note
This approach overlooks that $g(x)$ [and therefore $f(x)$] are concave (down) functions. This therefore suggests that $\log_{10}(e) > (0.434).$ Unfortunately, rounding error is relevant here, so the estimate of $\log_{10}(e) \approx (0.434)$ is probably best left alone.


$\underline{\text{Step 4:}}$

You have that $7^2 = (50 - 1) \implies 7^4 = (50)^2 - 2(50) + 1 = 2401.$

Examining $f(2401) - f(2400)$, note that $\dfrac{1}{2400} \times \dfrac{3}{7} = \dfrac{3}{16800} < \dfrac{1}{5000}.$

Therefore, $\log_{10}\left(7^4\right)$ may reasonably be estimated as $\log_{10}(2400) = (0.903) + (0.477) + (2) = (3.380) = (3.400 - 0.020).$

Therefore $\log_{10}(7) \approx \dfrac{1}{4} \times \dfrac{3400 - 20}{1000} = \dfrac{850 - 5}{1000} = (0.845).$

.....

Now, you can use that $(7 \times 11 \times 13) = (1001).$

This implies that $\log_{10}(13) \approx \log_{10}(1001) - 0.845 - 1.0414.$

Since $\dfrac{3}{7} \times \dfrac{1}{1000} \approx (0.0004)$,
$\log_{10}(1001)$ may reasonably be approximated by $(3.0004)$.

This implies that $\log_{10}(13) \approx (3.0004 - 1.8864) = (1.114).$


$\underline{\text{Step 5:}}$

You now have $\log_{10}\left(e^4\right) \approx 4 \times 0.434 = 1.736.$

$\log_{10} (4 \times 13) \approx 0.602 + 1.114 = 1.716.$

Therefore $(e^4) > (4 \times 13).$

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