# Comparing $e^{4}-2$ and $50$ without calculator

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?

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

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

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

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