Which is greater? Part 2 - - With cube roots

[lookagain]

This is in regards to this thread:

Which is greater? A = (3)(2^{1/3}) or B = 1 + (2)(3^{1/3})?

The question is as follows: If A = 3 \cdot 2^{\dfrac{1}{3}}, and B = 1 + 2 \cdot 3^{\dfrac{1}{3}} Which is greater? Answer: Calculator is not accepted on the test. Thanks in advance. :)

www.freemathhelp.com

Jomo was fortunate with his method, because the cube of the left-hand side equals 54.

$\displaystyle A = \ \sqrt[3]{58} \ \ \ and \ \ \ B = \ 1 + 2\sqrt[3]{3}$

Without using a calculator or computer, show which is greater, A or B.

 

[Deleted member 4993]

This is in regards to this thread:

Which is greater? A = (3)(2^{1/3}) or B = 1 + (2)(3^{1/3})?

The question is as follows: If A = 3 \cdot 2^{\dfrac{1}{3}}, and B = 1 + 2 \cdot 3^{\dfrac{1}{3}} Which is greater? Answer: Calculator is not accepted on the test. Thanks in advance. :)

www.freemathhelp.com

Jomo was fortunate with his method, because the cube of the left-hand side equals 54.

$\displaystyle A = \ \sqrt[3]{58} \ \ \ and \ \ \ B = \ 1 + 2\sqrt[3]{3}$

Without using a calculator or computer, show which is greater, A or B.

Can I use slide rule? I have one!

 

[lookagain]

Can I use slide rule? I have one!

You can use your slide rule to check the answer if you would like.

 

[ksdhart2]

$\displaystyle A = \ \sqrt[3]{58} \ \ \ and \ \ \ B = \ 1 + 2\sqrt[3]{3}$

Without using a calculator or computer, show which is greater, A or B.

Here's how I would do it:

$\displaystyle A = \sqrt[3]{58} = 58^{\frac{1}{3}} \implies \ln(A) = \frac{1}{3} \cdot \ln(58)$

$\displaystyle B - 1 = 2\sqrt[3]{3} = 2 \cdot 3^{\frac{1}{3}} \implies \ln(B - 1) = \ln(2) + \frac{1}{3} \cdot \ln(3)$

$\displaystyle A - (B - 1) = \frac{1}{3} \cdot \ln(58) - \frac{1}{3} \cdot \ln(3) - \ln(2) = \frac{1}{3} \cdot \left[\ln(58) - \ln(3) \right] - \ln(2) = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - \ln(2)$

$\displaystyle A - B = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - 1 - \ln(2) = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - \ln(2e)$

Now I'll use some crude approximations and known facts to bound these numbers.

$\displaystyle (e > 2) \wedge (2 > 1) \implies e < 2e < e^2 \implies 1 < \ln(2e) < 2 \implies 3 < \ln(2e) < 6$

$\displaystyle e \approx 2.7 \implies e^3 \approx 2.7^3 = 19.683$

$\displaystyle \frac{58}{3} = 19.\overline{33} \implies \frac{58}{3} < e^3$

The natural log function is monotonically increasing for $x > 0$ (if desired this can be a calculator-free subproof involving the derivative) so that means:

$\displaystyle \frac{58}{3} < e^3 \implies \ln \left( \frac{58}{3} \right) < \ln(e^3) \implies \ln \left( \frac{58}{3} \right) < 3$

From this I can conclude that:

$\displaystyle \ln \left( \frac{58}{3} \right) < 3\ln(2e) \implies \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) < \ln(2e)$

Both $\frac{58}{3}$ and $2e$ are greater than 1, so their natural logs must be positive. Putting everything together, we have:

$\displaystyle A - B = (\text{A positive number}) - (\text{A bigger positive number}) < 0$

And thus, $B$ must be the bigger of the two numbers.

 

[Steven G]

This is in regards to this thread:

Which is greater? A = (3)(2^{1/3}) or B = 1 + (2)(3^{1/3})?

The question is as follows: If A = 3 \cdot 2^{\dfrac{1}{3}}, and B = 1 + 2 \cdot 3^{\dfrac{1}{3}} Which is greater? Answer: Calculator is not accepted on the test. Thanks in advance. :)

www.freemathhelp.com

Jomo was fortunate with his method, because the cube of the left-hand side equals 54.

$\displaystyle A = \ \sqrt[3]{58} \ \ \ and \ \ \ B = \ 1 + 2\sqrt[3]{3}$

Without using a calculator or computer, show which is greater, A or B.

Yes, I was very lucky. Now understand that this was a test question that has to be done within a time limit so I think that my method *had to work*, otherwise the problem was unfair.

Now your question is entirely different and will take some thoughtl
Thanks for posting it.

 

[Steven G]

Here is my solution and I did not use a slide ruler although I did use a calculator to find 3rt(3) which I could have approximated by hand

 

[lookagain]

Here's how I would do it:

$\displaystyle A = \sqrt[3]{58} = 58^{\frac{1}{3}} \implies \ln(A) = \frac{1}{3} \cdot \ln(58)$

$\displaystyle B - 1 = 2\sqrt[3]{3} = 2 \cdot 3^{\frac{1}{3}} \implies \ln(B - 1) = \ln(2) + \frac{1}{3} \cdot \ln(3)$

$\displaystyle A - (B - 1) = \frac{1}{3} \cdot \ln(58) - \frac{1}{3} \cdot \ln(3) - \ln(2) = \frac{1}{3} \cdot \left[\ln(58) - \ln(3) \right] - \ln(2) = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - \ln(2)$

$\displaystyle A - B = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - 1 - \ln(2) = \frac{1}{3} \cdot \ln \left( \frac{58}{3} \right) - \ln(2e)$

I did not read beyond these lines, because of the far left ends of lines three and four.
It would make sense if line three began "ln(A) - ln(B - 1)," and if line four began
"ln(A) - ln(B)."

Because of editing time restrictions for that post of yours, perhaps if you were to copy-and-paste your
whole post and make amendments, and then post that, I could read that through.

 

[lookagain]

Here is my solution and I did not use a slide ruler although I did use a calculator to find 3rt(3) which I could have approximated by hand

You missed an expected opportunity, after the fifth line, to divide each side by 3,
before cubing each side, for instance.

 

[ksdhart2]

...It would make sense if line three began "ln(A) - ln(B - 1)," and if line four began
"ln(A) - ln(B)."

Augh! You're correct. This means that nothing I did works out. I got the correct answer, but only purely by coincidence. I'll give it some more thought and maybe something will come to me, but I don't see any way of correcting that. I can't say anything meaningful about $A - B$ based on $\ln(A) - \ln(B - 1) - 1$