Algebra Question of The Day -7

BigBeachBanana

BigBeachBanana

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Without the use of a calculator, which is larger? [imath]\large \log_{2}3 [/imath] or [imath]\large \log_{3}5?[/imath]
Justify your answer (in spoiler). :)
 
[math]9>8\Leftrightarrow\log_29>\log_28\Leftrightarrow 2\log_23>3\Leftrightarrow\log_23>\frac{3}{2}[/math][math]25<27\Leftrightarrow\log_3{25}<\log_3{27}\Leftrightarrow 2\log_35<3\Leftrightarrow\log_35<\frac{3}{2}[/math]So [math]\log_23>\log_35[/math]
 
This is one of those coffin problems which I have seen before so I will not accept credit for solving it.
Note that the two given values are between 1 and 2. Hmm, maybe one of those numbers is less than 3/2 and one of them is more than 3/2. It can easily be shown that log_2(3)> 3/2 while log_3(5) < 3/2. ...[\spoiler]
 
Steven G said:
This is one of those coffin problems which I have seen before so I will not accept credit for solving it.
Note that the two given values are between 1 and 2. Hmm, maybe one of those numbers is less than 3/2 and one of them is more than 3/2. It can easily be shown that log_2(3)> 3/2 while log_3(5) < 3/2. ...[\spoiler]
Click to expand...
That's okay. We wouldn't give you credit, anyway. :)

-Dan
 
This may be a sloppy sketch. Not sure if it can be made more crisp.

EDIT: Well I seem to have made an error. Can’t find it though?

[math] x = \log_3(5) \implies x > 1 > 0 \ \because \ 5 > 3.\\ y = \log_2(3) \implies y > 1 > 0 \ \because \ 3 > 2.\\ y = \log_2(3) \implies 2^y = 3 \implies \\ y \log_3(2) = \log_3(3) = 1 \implies \dfrac{1}{y} = \log_3(2).\\ \text {But } \log_3(2) > 0 \ \because \ 3 > 1 \text { and } 2 > 1.\\ \therefore \ \log_3(5) * \log_3(2) = \log_3 \left ( 5^{\log_3(2)} \right ) > \log_3 \left ( 5^0 \right ) = 1 \implies \\ \dfrac{x}{y} = x * \dfrac{1}{y} = \log_3(5) * log_3(2) > 1 \implies \dfrac{x}{y} > 1 \implies \\ x > y \implies \log_3(5) > \log_2(3). [/math]
 
Last edited:
Corner time!
\(\displaystyle \log_3 {5^0} = \log_3 {1} = 1??\) I don't think so.
 
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