logarithm: 1/4 >1/8 leads (by logs) to 2 > 3; what is going wrong here?

[shahar]

Where is/are the error/s in the picture?

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\(\displaystyle \dfrac{1}{4}\, >\, \dfrac{1}{8}\)

\(\displaystyle \left(\dfrac{1}{2}\right)^2\, >\, \left(\dfrac{1}{2}\right)^3\)

\(\displaystyle 2\,\log_{10}\left(\dfrac{1}{2}\right)\, >\, 3\, \log_{10}\left(\dfrac{1}{2}\right)\)

\(\displaystyle 2\, >\, 3\)

 

[sinx]

Where is/are the error/s in the picture?

I don't think you can write log₁₀ (1/2)² =2log₁₀ (1/2)
however, log₁₀ (1/2)² =2log₁₀1-2log₁₀2
similarly, log₁₀ (1/2)³ =3log₁₀1-3log₁₀2

following this your inequality is
-2>-3

 

[pka]

Where is/are the error/s in the picture?

The number \(\displaystyle \log_{10}\left(\frac{1}{2}\right)<0\). So you have divided by a negative.

 

[Dr.Peterson]

Where is/are the error/s in the picture?

Everything is valid until the last line:

1/4 > 1/8
(1/2)^2 > (1/2)^3
2log₁₀(1/2) > 3log₁₀(1/2)

But on the last line, you have divided both sides by log₁₀(1/2), forgetting that this is a negative number, and therefore changes the sense of the inequality. The last line should be (as you know)

2 < 3

 

[JeffM]

Earlier today, I posted a response that had a serious technical error. Consequently, I deleted the response. But there was a substantive point to that response so I am posting this revision.

It is true, although the initial post did not justify it, that

\(\displaystyle 0 < a < b \implies log(a) < log(b)\)

for any valid base for a logarithm. This is not true for every function. It is NOT generally true that

\(\displaystyle 0 < a < b \implies f(a) < f(b).\)

That is true only for monotonically increasing functions.

Furthermore, this logic has no flaw:

\(\displaystyle \dfrac{1}{4} > \dfrac{1}{8} > 0 \\

\dfrac{1}{4} = \left ( \dfrac{1}{2} \right )^2. \\

\dfrac{1}{8} = \left ( \dfrac{1}{2} \right )^3. \\

\therefore \left ( \dfrac{1}{2} \right )^2 > \left ( \dfrac{1}{2} \right )^3 \implies

2 log \left \{ \left ( \dfrac{1}{2} \right )^2 \right \} > 3 log \left \{ \left ( \dfrac{1}{2} \right )^2 \right \}.\)

But the very same theorem that justifies the prior line also justifies

\(\displaystyle 0 < \dfrac{1}{2} < 0 \implies log \left ( \dfrac{1}{2} \right ) < log(1) \implies log \left ( \dfrac{1}{2} \right ) < 0.\)

In other words, the reasoning originally cited uses an unstated theorem and then implicitly contradicts that same theorem by implying that

\(\displaystyle log \left ( \dfrac{1}{2} \right ) > 0.\)

 

[HallsofIvy]

Earlier today, I posted a response that had a serious technical error. Consequently, I deleted the response. But there was a substantive point to that response so I am posting this revision.

It is true, although the initial post did not justify it, that

\(\displaystyle 0 < a < b \implies log(a) < log(b)\)

for any valid base for a logarithm.

This is not correct. A number between 0 and 1 is a valid base but the logarithm to such a base is a decreasing function. a< b implies log(a)< log(b) only for base greater than 1.

This is not true for every function. It is NOT generally true that

\(\displaystyle 0 < a < b \implies f(a) < f(b).\)

That is true only for monotonically increasing functions.

Furthermore, this logic has no flaw:

\(\displaystyle \dfrac{1}{4} > \dfrac{1}{8} > 0 \\

\dfrac{1}{4} = \left ( \dfrac{1}{2} \right )^2. \\

\dfrac{1}{8} = \left ( \dfrac{1}{2} \right )^3. \\

\therefore \left ( \dfrac{1}{2} \right )^2 > \left ( \dfrac{1}{2} \right )^3 \implies

2 log \left \{ \left ( \dfrac{1}{2} \right )^2 \right \} > 3 log \left \{ \left ( \dfrac{1}{2} \right )^2 \right \}.\)

But the very same theorem that justifies the prior line also justifies

\(\displaystyle 0 < \dfrac{1}{2} < 0 \implies log \left ( \dfrac{1}{2} \right ) < log(1) \implies log \left ( \dfrac{1}{2} \right ) < 0.\)

In other words, the reasoning originally cited uses an unstated theorem and then implicitly contradicts that same theorem by implying that

\(\displaystyle log \left ( \dfrac{1}{2} \right ) > 0.\)

 

[JeffM]

This is not correct. A number between 0 and 1 is a valid base but the logarithm to such a base is a decreasing function. a< b implies log(a)< log(b) only for base greater than 1.

Right you are.