Not really sure with what u got. Please do help

[MethMath11]

Looks fun.

 

[MethMath11]

With what i got' sorry for the typo

 

[Dr.Peterson]

Your notation is unfamiliar to me (and I've seen a lot!). Is the superscript on the left of "log" meant to indicate the base of the log, or is that the argument and the base is written after "log"? It may help if you can give us a link to a page that uses your notation, or an image from your textbook that defines it.

One way or another, I would start with the change of base formula, probably in terms of base 2 (since 8 and 32 are powers of 2).

 

[MethMath11]

Your notation is unfamiliar to me (and I've seen a lot!). Is the superscript on the left of "log" meant to indicate the base of the log, or is that the argument and the base is written after "log"? It may help if you can give us a link to a page that uses your notation, or an image from your textbook that defines it.

One way or another, I would start with the change of base formula, probably in terms of base 2 (since 8 and 32 are powers of 2).

 

[Dr.Peterson]

You didn't actually answer my question about the meaning of the notation, but the answers are consistent with my guess that the first number is the base. So in the notation I am familiar with, the problem is:

If [MATH]\log_{3-\sqrt{7}}32 = a[/MATH], then find [MATH]\log_8(3+\sqrt{7})[/MATH],​

where the subscripts are the bases.

As I suggested, I would express the given log in terms of base 2, and solve for [MATH]\log_2(3-\sqrt{7})[/MATH]. Do you know the change of base, formula, [MATH]\log_a x = \frac{\log_b x}{\log_b a}[/MATH] ? (In your notation, [MATH]^a\log x = \frac{^b\log x}{^b\log a}[/MATH])

Also, one way to deal with the conjugates [MATH]3-\sqrt{7}[/MATH] and [MATH]3+\sqrt{7}[/MATH] is to note that their product is 2, so [MATH]3+\sqrt{7} = \frac{2}{ 3-\sqrt{7}}[/MATH].

Please show what you can do, so we can help you wherever you are having trouble.

It is fun!

 

[MethMath11]

You didn't actually answer my question about the meaning of the notation, but the answers are consistent with my guess that the first number is the base. So in the notation I am familiar with, the problem is:

If [MATH]\log_{3-\sqrt{7}}32 = a[/MATH], then find [MATH]\log_8(3+\sqrt{7})[/MATH],​

where the subscripts are the bases.

As I suggested, I would express the given log in terms of base 2, and solve for [MATH]\log_2(3-\sqrt{7})[/MATH]. Do you know the change of base, formula, [MATH]\log_a x = \frac{\log_b x}{\log_b a}[/MATH] ? (In your notation, [MATH]^a\log x = \frac{^b\log x}{^b\log a}[/MATH])

Also, one way to deal with the conjugates [MATH]3-\sqrt{7}[/MATH] and [MATH]3+\sqrt{7}[/MATH] is to note that their product is 2, so [MATH]3+\sqrt{7} = \frac{2}{ 3-\sqrt{7}}[/MATH].

Please show what you can do, so we can help you wherever you are having trouble.

It is fun!

I really thought it was my writing, turns out not everyone uses the same notation as we did. I really should've learn more. Thank you

 

[MethMath11]

I really thought it was my writing, turns out not everyone uses the same notation as we did. I really should've learn more. Thank you

And thank you for showing me how to deal with the conjungates. Never thought of it that way

 

[Dr.Peterson]

Sounds like you have succeeded. Great!

Note that there are many variations in notation around the world, and the notation you are taught is correct for you. (There may be good reasons for it, and perhaps it will spread!) But clearly if a relatively rare notation is used in your classes, it can make it difficult to find help online, so it would be helpful if instructors at least mentioned other notations.

People often imagine that math is a "universal language"; but that language has many dialects!