Find geometric sequence terms (w/ logs): log_4(9), log_2(a), log_5(2) * log_9(25)

[dr.trovacek]

Hello, I have this assignment that I have problem with. Assignment is from a 4th grade high school book - at least in my country. I'll try to translate it as accurate as I can, but please consider that I do not know what exact words you would use for the same assignment in English.

For wich values of a real number a, numbers \(\displaystyle \log_4{9}, \log_2{a},\log_5{2} \cdot \log_9{25} \) (in that order) are consecutive terms of a geometric sequence.


I used the formula for geometric mean: \(\displaystyle a_n= \sqrt{a_{n-1} \cdot a_{n+1}} \) (1)
where \(\displaystyle \log_2{a} \) is the middle term \(\displaystyle a_n\)

I used properties of logarithms to determine the following:

\(\displaystyle \log_4 {9}=\log_2{3}= \log_2{3}\)

\(\displaystyle \log_5{2} \cdot \log_3{5}= \frac{\log_2{2}}{\log_2{5}} \cdot \frac{\log_2{5}}{\log_2{3}}= \frac{1}{\log_2{3}}\)

So now, using the above mentioned formula (1), we get the condition:

\(\displaystyle \log_2{a} = \sqrt { \log_2{3} \cdot \frac{1}{\log_2{3}} } = \sqrt{1}\)

Now I would say that only solution to this is \(\displaystyle \log_2{a} =1\) that is \(\displaystyle a = 2^1 \)

So we get a geometric sequence: \(\displaystyle \log_2{3}, 1,\frac{1}{\log_2{3}} \)


However, the solution given in the textbook is a little different.

They say that the condition is given like this:

\(\displaystyle \log^2_2{a} =\log_2{3} \cdot \frac{1}{\log_2{3}} = 1 \)

Not they take the sqare root of this expression and to get:

\(\displaystyle \log_2{a} = \pm 1 \) that is \(\displaystyle a = 2 \) or \(\displaystyle a = \frac{1}{2} \)

So we now have a sequence: \(\displaystyle \log_2{3}, 1,\frac{1}{\log_2{3}} \) and \(\displaystyle \log_2{3}, -1,\frac{1}{\log_2{3}} \)


In conclusion, both solutions kinda make sense to me when you look at how they were derived.
On the other hand, the solution where the mid term is \(\displaystyle -1 \) is not valid in the domain of natural numbers if we look at the condition for the geometric sequence (1), since we would get:

\(\displaystyle -1 = \sqrt { \log_2{3} \cdot \frac{1}{\log_2{3}} } \)

This ought to be a complex number.

So this is my take on the problem. Since I'm not 100% certain, because maybe I'm missing some logarithm property or something, any advice would be much appreciated. Thank you!

 

[tkhunny]

Hello, I have this assignment that I have problem with. Assignment is from a 4th grade high school book - at least in my country. I'll try to translate it as accurate as I can, but please consider that I do not know what exact words you would use for the same assignment in English.

For wich values of a real number a, numbers \(\displaystyle \log_4{9}, \log_2{a},\log_5{2} \cdot \log_9{25} \) (in that order) are consecutive terms of a geometric sequence.


I used the formula for geometric mean: \(\displaystyle a_n= \sqrt{a_{n-1} \cdot a_{n+1}} \) (1)
where \(\displaystyle \log_2{a} \) is the middle term \(\displaystyle a_n\)

I used properties of logarithms to determine the following:

\(\displaystyle \log_4 {9}=\log_2{3}= \log_2{3}\)

\(\displaystyle \log_5{2} \cdot \log_3{5}= \frac{\log_2{2}}{\log_2{5}} \cdot \frac{\log_2{5}}{\log_2{3}}= \frac{1}{\log_2{3}}\)

So now, using the above mentioned formula (1), we get the condition:

\(\displaystyle \log_2{a} = \sqrt { \log_2{3} \cdot \frac{1}{\log_2{3}} } = \sqrt{1}\)

Now I would say that only solution to this is \(\displaystyle \log_2{a} =1\) that is \(\displaystyle a = 2^1 \)

So we get a geometric sequence: \(\displaystyle \log_2{3}, 1,\frac{1}{\log_2{3}} \)


However, the solution given in the textbook is a little different.

They say that the condition is given like this:

\(\displaystyle \log^2_2{a} =\log_2{3} \cdot \frac{1}{\log_2{3}} = 1 \)

Not they take the sqare root of this expression and to get:

\(\displaystyle \log_2{a} = \pm 1 \) that is \(\displaystyle a = 2 \) or \(\displaystyle a = \frac{1}{2} \)

So we now have a sequence: \(\displaystyle \log_2{3}, 1,\frac{1}{\log_2{3}} \) and \(\displaystyle \log_2{3}, -1,\frac{1}{\log_2{3}} \)


In conclusion, both solutions kinda make sense to me when you look at how they were derived.
On the other hand, the solution where the mid term is \(\displaystyle -1 \) is not valid in the domain of natural numbers if we look at the condition for the geometric sequence (1), since we would get:

\(\displaystyle -1 = \sqrt { \log_2{3} \cdot \frac{1}{\log_2{3}} } \)

This ought to be a complex number.

So this is my take on the problem. Since I'm not 100% certain, because maybe I'm missing some logarithm property or something, any advice would be much appreciated. Thank you!

No reason that shouldn't have worked, excepting the confusion. Did we start with log(5) or log(25)?

 

[Dr.Peterson]

In conclusion, both solutions kinda make sense to me when you look at how they were derived.
On the other hand, the solution where the mid term is \(\displaystyle -1 \) is not valid in the domain of natural numbers if we look at the condition for the geometric sequence (1), since we would get:

\(\displaystyle -1 = \sqrt { \log_2{3} \cdot \frac{1}{\log_2{3}} } \)

This ought to be a complex number.

First, why do you mention natural numbers (positive integers)? Do you just mean positive numbers? Is that a part of the definition you are using for a geometric sequence?

Second, the negative square root issue is unrelated to complex numbers (which relate to square roots of negative numbers), but to the fact that any positive number has two square roots, only one of which is denoted by the radical symbol.

So your error (relative to the book's assumptions) is in using that radical, which implies that geometric means are always positive. Some authors may define it that way, but clearly not this one. Or rather, the problem isn't about geometric means, but about a geometric sequence, whose common ratio can certainly be either positive or negative. Whatever one says about negative geometric means, the middle term can be negative.

The proper statement would be that a_n² = a_n-1 a_n+1. This allows for either sign.

 

[dr.trovacek]

First, why do you mention natural numbers (positive integers)? Do you just mean positive numbers? Is that a part of the definition you are using for a geometric sequence?

Second, the negative square root issue is unrelated to complex numbers (which relate to square roots of negative numbers), but to the fact that any positive number has two square roots, only one of which is denoted by the radical symbol.

So your error (relative to the book's assumptions) is in using that radical, which implies that geometric means are always positive. Some authors may define it that way, but clearly not this one. Or rather, the problem isn't about geometric means, but about a geometric sequence, whose common ratio can certainly be either positive or negative. Whatever one says about negative geometric means, the middle term can be negative.

The proper statement would be that a_n² = a_n-1 a_n+1. This allows for either sign.

Sorry, I used the wrong term, I didn't mean natural numbers, but real numbers.

Yeah I think I can see how I "flipped" the property of complex numbers, that is misinterpreted it in this case.

I used that statement with the radical because that is the formula In my textbook. But yeah I get that the middle term of a geometric sequence could allow for middle terms to be negative numbers.

Thank you!