Log Question 2: electric boogaloo

[Almonds]

I've spent the last ~ hour and a half on this, I've attempted to change the bases to both log base 2 and log base 5 but I just end up with gibberish (at least to me). I really don't want a flat out answer but I feel like I'm missing something important and just don't know what it is.

Thanks to anyone who stops to help.

[MATH]\log _2\left(2\right)x+5\log _3\left(x\right)\cdot \log _4\left(x\right)+\log _5\left(2\right)x[/MATH]
(Link to the original question in case I'm reading it wrong: https://prnt.sc/wbyl88)

 

[Almonds]

Ok at this point I just really want to know what x is equal to, I've shoved it in math-way and symbol lab and neither can give me an answer.

 

[Harry_the_cat]

By just looking at it x=1 is a solution but possibly not the only one.

 

[Almonds]

By just looking at it x=1 is a solution but possibly not the only one.

When I sub 1 in I get 1+ log 2/ log 5.

Am I meant to make a quadratic out of it?

 

[lev888]

When I sub 1 in I get 1+ log 2/ log 5.

Am I meant to make a quadratic out of it?

When x=1 all terms are 0. You didn't write the equation correctly.

 

[Almonds]

When x=1 all terms are 0. You didn't write the equation correctly.

log2(2)x+5log3(x)⋅log4(x)+log5(2)x
-->
log2(2)(1)+5log3(1)⋅log4(1)+log5(2)(1)
-->
1+0⋅0+log5(2)
-->
1+log5(2)
No?

If that's not how it's meant to be written could you show me, cause I'm just lost.

 

[JeffM]

I just this second started looking at this, but both harry and lev already provided a correct answer

[MATH]f(x) = log_2^2(x) + 5 log_3(x) * log_4(x) + log_5^2(x).[/MATH]
The log of 1, no matter what the base, is 0.

[MATH](0)^2 + 5(0) * 0 + (0)^2 = 0.[/MATH]
[MATH]\therefore f(1) = 0.[/MATH]
This problem uses a notation for logarithms that I hate. x is not a factor; it is an argument. 2 is not an argument but a base. Parentheses help.

 

[Almonds]

I just this second started looking at this, but both harry and lev already provided a correct answer

[MATH]f(x) = log_2^2(x) + 5 log_3(x) * log_4(x) + log_5^2(x).[/MATH]
The log of 1, no matter what the base, is 0.

[MATH](0)^2 + 5(0) * 0 + (0)^2 = 0.[/MATH]
[MATH]\therefore f(1) = 0.[/MATH]
I hate the notation used for logarithms (without parentheses) in that problem. x is not a factor; it is an argument. 2 is not an argument but a base.

Could you elaborate a little? These question were given to me by a tutor and are not from my curriculum, I've never heard of x being an argument or even what an argument is within a maths context. How do I 'deal' with/ treat them? Sorry if this is a big ask but I have no context here.

 

[lev888]

Could you elaborate a little? These question were given to me by a tutor and are not from my curriculum, I've never heard of x being an argument or even what an argument is within a maths context. How do I 'deal' with/ treat them? Sorry if this is a big ask but I have no context here.

Argument of a function is the variable that gets assigned 'input' values so that we can calculate the result.
Seems strange that a tutor who is aware of where you are in the math curriculum would give you this problem.

 

[JeffM]

You are, I presume, familiar with functions.

z = f(x, y) or u = g(v). In these examples, z and u are the results of the functions f and g respectively. And x and y are the arguments of f while v is the argument of g. If you are used to the machine analogy used to describe functions, arguments are inputs, and results are outputs.

The notation for logarithms has always been a little odd. They are really a family of related functions. A standard notation in modern form would look like this

[MATH]log(b, \ x)[/MATH]
where b is the base and x is the variable of primary interest.

Originally, however, logs were primarily used to assist in computation and always were to the base 10. Because the base was always the same, it was easier to just write log(x) with the 10 assumed. Later, with the development of calculus, logs to the base e became important. They were written ln(x) to distinguish them from logs with the base 10. The original use of logarithms to assist in arithmetic has been lost with the pocket calculator, but the log to the base e is still very important. So some people now write log(x) to mean a logarithm to the base e, and others still write log(x) to mean a logarithm to the base 10. Moreover, it is standard to use a subscript to represent the base. Moreover, many people omit the parentheses around functions indicated by abbreviations like tan or log. The notation is confusing enough without omitting parentheses.

 

[Almonds]

Argument of a function is the variable that gets assigned 'input' values so that we can calculate the result.
Seems strange that a tutor who is aware of where you are in the math curriculum would give you this problem.

Truth be told I think he either misread the question when assigning it or forgot to explain it before we went on break.

 

[JeffM]

Finally, I have to say that this is a trick question. I doubt there is an analytic way to deal with it. But whenever you see 0 as a result in something involving logarithms, you should consider whether the argument is 1 because the logarithm of 1, NO MATTER WHAT THE BASE, is always 0. It is a problem about recognizing a fundamental commonality about logarithms rather than about a technique for solving equations.

 

[Almonds]

You are, I presume, familiar with functions.

z = f(x, y) or u = g(v). In these examples, z and u are the results of the functions f and g respectively. And x and y are the arguments of f while v is the argument of g. If you are used to the machine analogy used to describe functions, arguments are inputs, and results are outputs.

The notation for logarithms has always been a little odd. They are really a family of related functions. A standard notation in modern form would look like this

[MATH]log(b, \ x)[/MATH]
where b is the base and x is the variable of primary interest.

Originally, however, logs were primarily used to assist in computation and always were to the base 10. Because the base was always the same, it was easier to just write log(x) with the 10 assumed. Later, with the development of calculus, logs to the base e became important. They were written ln(x) to distinguish them from logs with the base 10. The original use of logarithms to assist in arithmetic has been lost with the pocket calculator, but the log to the base e is still very important. So some people now write log(x) to mean a logarithm to the base e, and others still write log(x) to mean a logarithm to the base 10. Moreover, it is standard to use a subscript to represent the base. Moreover, many people omit the parentheses around functions indicated by abbreviations like tan or log. The notation is confusing enough without omitting parentheses.

Thanks for the reply, but I am not that familiar with functions. I think I ought to consult my tutor about this and receive a more personalised explanation. Thanks for taking the time to answer though, much appreciated.

 

[HallsofIvy]

Frankly, your original post makes no sense! You have an expression, \(\displaystyle log_2(x)+ 5log_3(x)log_4(x)- log_5(2)x\), NOT an equation so you cannot "solve" for x.
If it is the different bases of the logarithms then you can use the fact that \(\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}\) where "b" is any positive number. Writing "log(x)" to mean logarithm base 10 (or any other positive number), \(\displaystyle log_2(x)+ 5log_3(x)log_4(x)- log_5(2)x= \frac{log(x)}{log(2)}+ \frac{5 log(x)}{log(3)}- \frac{log(2)x}{log(5)}\)

 

[Dr.Peterson]

I've spent the last ~ hour and a half on this, I've attempted to change the bases to both log base 2 and log base 5 but I just end up with gibberish (at least to me). I really don't want a flat out answer but I feel like I'm missing something important and just don't know what it is.

Thanks to anyone who stops to help.

[MATH]\log _2\left(2\right)x+5\log _3\left(x\right)\cdot \log _4\left(x\right)+\log _5\left(2\right)x[/MATH]
(Link to the original question in case I'm reading it wrong: https://prnt.sc/wbyl88)

First, the actual equation is [MATH]\log_2^2(x) + 5 \log_3(x) \log_4(x) + \log_5^2(x) = 0[/MATH]. It's really important to type things correctly; thanks for including the image, which you could have just pasted in:



The way I would start to solve this algebraically would be, as you suggest, to use the change of base formula. It would really help to see that work, because it would have shown us more quickly what you do and do not understand.

Hopefully you got something like this:

[MATH]\left(\frac{\ln x}{\ln 2}\right)^2 + 5\left(\frac{\ln x}{\ln 3}\right)\left(\frac{\ln x}{\ln 4}\right) + \left(\frac{\ln x}{\ln 5}\right)^2 = 0[/MATH]
With a little thought, you can factor out [MATH](\ln x)^2[/MATH], and will realize that the other factor is just a non-zero constant, which you can ignore. Then you just have to solve [MATH](\ln x)^2 = 0[/MATH], which is easy enough. All the extra junk is just there to make it look worse than it is, as others have pointed out.

It may be that the factoring out is the step you missed.

 

[JeffM]

First, the actual equation is [MATH]\log_2^2(x) + 5 \log_3(x) \log_4(x) + \log_5^2(x) = 0[/MATH]. It's really important to type things correctly; thanks for including the image, which you could have just pasted in:

View attachment 24073

The way I would start to solve this algebraically would be, as you suggest, to use the change of base formula. It would really help to see that work, because it would have shown us more quickly what you do and do not understand.

Hopefully you got something like this:

[MATH]\left(\frac{\ln x}{\ln 2}\right)^2 + 5\left(\frac{\ln x}{\ln 3}\right)\left(\frac{\ln x}{\ln 4}\right) + \left(\frac{\ln x}{\ln 5}\right)^2 = 0[/MATH]
With a little thought, you can factor out [MATH](\ln x)^2[/MATH], and will realize that the other factor is just a non-zero constant, which you can ignore. Then you just have to solve [MATH](\ln x)^2 = 0[/MATH], which is easy enough. All the extra junk is just there to make it look worse than it is, as others have pointed out.

It may be that the factoring out is the step you missed.

Very very nice. It shows that there is a unique answer.

 

[Almonds]

First, the actual equation is [MATH]\log_2^2(x) + 5 \log_3(x) \log_4(x) + \log_5^2(x) = 0[/MATH]. It's really important to type things correctly; thanks for including the image, which you could have just pasted in:

View attachment 24073

The way I would start to solve this algebraically would be, as you suggest, to use the change of base formula. It would really help to see that work, because it would have shown us more quickly what you do and do not understand.

Hopefully you got something like this:

[MATH]\left(\frac{\ln x}{\ln 2}\right)^2 + 5\left(\frac{\ln x}{\ln 3}\right)\left(\frac{\ln x}{\ln 4}\right) + \left(\frac{\ln x}{\ln 5}\right)^2 = 0[/MATH]
With a little thought, you can factor out [MATH](\ln x)^2[/MATH], and will realize that the other factor is just a non-zero constant, which you can ignore. Then you just have to solve [MATH](\ln x)^2 = 0[/MATH], which is easy enough. All the extra junk is just there to make it look worse than it is, as others have pointed out.

It may be that the factoring out is the step you missed.

I haven't actually used/ learned about natural logs as of yet and the reason I miss wrote the equation is because I've never seen this kind of notation before. I think the question as a whole is just a bit outside my skill level and will probably need some more practice before I have a run at it. But just for conversation, how did you change natural log bases for all of them? The only reason I ask is because when I changed bases to base 2 I also got consecutive denominators (2,3,4,5) and wanted to know if that was connected in someway.

 

[Steven G]

First of all you never wrote an equation. An equation has an equal sign. Seriously how did you expect us to know what you wrote was to equal 0? Why not 14 or log₃x?

There is something called the change of base formula for logs. If you have [math]\log_ab[/math] you can pick any base, say c, and it will be true that [math]\log_ab=\dfrac{\log_cb}{\log_ca}[/math]
For example if you needed to approximate [math]log_37[/math] on your calculator, since you do not have a base 3 button on your calculator, you can calculate [math]\dfrac{\log7}{\log3}\ or \ \dfrac{\ln7}{\ln3}[/math] as all three will give the same answer.

 

[Dr.Peterson]

I haven't actually used/ learned about natural logs as of yet and the reason I miss wrote the equation is because I've never seen this kind of notation before. I think the question as a whole is just a bit outside my skill level and will probably need some more practice before I have a run at it. But just for conversation, how did you change natural log bases for all of them? The only reason I ask is because when I changed bases to base 2 I also got consecutive denominators (2,3,4,5) and wanted to know if that was connected in someway.

How did YOU change bases? Please show us.

This discussion is difficult because all we seem to hear from you is what you don't know (which appears to be just about everything you need in order to even read this problem); so we don't know what we can use that you will understand.

Another problem is that what work you have shown is all in an unreadable notation. Please learn from us how to type logs properly, so we can understand you. To write "[MATH]\log_5^2 x[/MATH]", you can type "(log_5(x))^2". Do not type "log5(2)x". From posts #4 and #6, you seem to be misreading your own writing.

 

[Almonds]

How did YOU change bases? Please show us.

This discussion is difficult because all we seem to hear from you is what you don't know (which appears to be just about everything you need in order to even read this problem); so we don't know what we can use that you will understand.

Another problem is that what work you have shown is all in an unreadable notation. Please learn from us how to type logs properly, so we can understand you. To write "[MATH]\log_5^2 x[/MATH]", you can type "(log_5(x))^2". Do not type "log5(2)x". From posts #4 and #6, you seem to be misreading your own writing.

Alright, at this point I'm just gonna wait. I would show what I've done but I worked under the assumption that x was a factor and not an argument so what I have done is useless. I get where you're coming from but as I stated above I just don't know enough to do this.

Again, thanks for the help.