But we can't compare them!

Algebra Level 1

Which of the following logarithms is greater? log 9 71 or log 8 61 \color{#20A900}{\log _{9}{71}} ~~~~\text{or}~~~~\color{#3D99F6}{\log _{8}{61}}

log 9 71 \log_{9 }{71} log 8 61 \log_{8}{61}

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3 solutions

Note first that 61 64 > 71 81 . \dfrac{61}{64} \gt \dfrac{71}{81}.

(Proof: We have that 30 640 < 30 243 3 64 < 10 81 1 3 64 > 1 10 81 61 64 > 71 81 . \dfrac{30}{640} \lt \dfrac{30}{243} \Longrightarrow \dfrac{3}{64} \lt \dfrac{10}{81} \Longrightarrow 1 - \dfrac{3}{64} \gt 1 - \dfrac{10}{81} \Longrightarrow \dfrac{61}{64} \gt \dfrac{71}{81}. )

By the change of base rule we then have that

log 8 ( 61 64 ) > log 8 ( 71 81 ) = log 9 ( 71 81 ) log 9 ( 8 ) > log 9 ( 71 81 ) , \log_{8}\left(\dfrac{61}{64}\right) \gt \log_{8}\left(\dfrac{71}{81}\right) = \dfrac{\log_{9}\left(\dfrac{71}{81}\right)}{\log_{9}(8)} \gt \log_{9}\left(\dfrac{71}{81}\right),

since we also know that 0 < log 9 ( 8 ) < 1. 0 \lt \log_{9}(8) \lt 1. This implies that

( log 8 ( 61 ) log 8 ( 64 ) ) > ( log 9 ( 71 ) log 9 ( 81 ) ) log 8 ( 61 ) > log 9 ( 71 ) , (\log_{8}(61) - \log_{8}(64)) \gt (\log_{9}(71) - \log_{9}(81)) \Longrightarrow \boxed{\log_{8}(61) \gt \log_{9}(71)},

since log 8 ( 64 ) = log 9 ( 81 ) = 2. \log_{8}(64) = \log_{9}(81) = 2.

Whenever I see the word "clearly" in a solution, I think the writer has not bothered to write up an explanation. The point of a solution is to work through the problem so that it makes sense to anyone! "Clearly", "evidently", "obviously" are not, in my opinion, words used in a good solution.

A Former Brilliant Member - 5 years, 7 months ago

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Thanks for the scolding, Andrei. :) I do realize that the use of these words raise red flags for some, but in my opinion they can be used when writing for specific audiences, (although I do draw the line at the word "obviously"; that one is a definite non-starter in any situation). I felt that, for a level 3 problem solver, the fact that 61 64 > 71 81 \frac{61}{64} \gt \frac{71}{81} would be quite evident upon observation, hence I felt I could use the word "clearly" without appearing intimidating or condescending. However, out of respect for you and your criticism, I have edited my solution to include a brief proof of my initial claim. I hope that you find it suitable.

In considering how to phrase a solution, there is a fine line between offering too much exposition and too little. For some readers "too much" can be tedious and unnecessary, and for others "too little" can be cryptic and unsatisfying. I prefer brevity so long as it does not compromise clarity, and in this case I felt that my initial claim was sufficiently self-evident to justify the use of the word "clearly" without further explanation. That being said, I did find it an interesting exercise to provide as straightforward a proof as possible. :)

Brian Charlesworth - 5 years, 7 months ago

61 61 is closer enough to 8 2 8^{2} than 71 71 is to 9 2 9^{2} , so l o g 8 61 log_{8}61 is greater. That was my reasoning

While that's clever intuition that worked out here, it turns out that this is not always the case. For example, 55 is closer to 8 2 = 64 8^2=64 than 71 is to 9 2 = 81 , 9^2=81, but log 8 ( 55 ) < log 9 ( 71 ) . \log_{8}(55) < \log_{9}(71).

Eli Ross Staff - 5 years, 7 months ago

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Yes, we even have that log 8 ( 56 ) < log 9 ( 71 ) , \log_{8}(56) \lt \log_{9}(71), which would be quite risky to intuit.

Brian Charlesworth - 5 years, 7 months ago

This intuition is dependent on the ratio of given values. I am not a real analysis student, not currently, maybe in future I will do. But my mathematical experience tells me that analysing such things requires the analysis of their ratios.

Laxmi Narayan Bhandari Xth B - 8 months, 3 weeks ago
Harmony Luce
Mar 30, 2016

By change-of-base, we have

log 8 61 = log 61 log 8 = log 61 3 log 2 = 1 3 log 2 61 \log_8 61 = \dfrac{\log 61}{\log 8} = \dfrac{\log 61}{3\log 2} = \dfrac{1}{3}\log_2 61

and

log 9 71 = log 71 log 9 = log 71 2 log 3 = 1 2 log 3 71 \log_9 71 = \dfrac{\log 71}{\log 9} = \dfrac{\log 71}{2\log 3} = \dfrac{1}{2}\log_3 71 .

Note: Both log 2 61 \log_2 61 and log 3 71 \log_3 71 are positive.

Observe that,

log 2 61 < 6 1 3 log 2 61 < 2 \log_2 61 < 6 \implies \dfrac{1}{3}\log_2 61 < 2

and

log 3 71 < 4 1 2 log 3 71 < 2 \log_3 71 < 4 \implies \dfrac{1}{2}\log_3 71 < 2 .

Notice that it only takes one-third of log 2 61 \log_2 61 to be less than 2, whereas it takes one-half of log 3 71 \log_3 71 to be less than 2. Since both log 2 61 \log_2 61 and log 3 71 \log_3 71 are positive, it must follow that log 2 61 > log 3 71 \log_2 61 > \log_3 71 .

Equivalently, log 8 61 > log 9 71 \log_8 61 > \log_9 71 .

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