Solving Log Equations: A Couple of Tricky Questions | Homework Help

[MadmanMurray]

Homework Statement

1.) 3logx²y + 2logxy
2.) 4logabc - 2loga²b - 3logbc

Homework Equations

The Attempt at a Solution

I know that 3logx² is the same as 6logx but I don't know what to do since there's a y there

 

[snipez90]

What is the question? If it's to simplify as much as possible, you can use log(xy) = log(x) + log(y).

 

[MadmanMurray]

Yep that's the question. I'm just confused about what to do with all these coefficients.
Is 3logx²y the same as logx⁶y³??

Can I write logx⁶y³ as logx^{6[/sup + log]y3 or do I have to get rid of them powers first?}

 

[NoMoreExams]

Yes assuming x^2y are both an argument of your log

 

[tiny-tim]

Can I write logx⁶y³ as logx⁶ + logy³ or do I have to get rid of them powers first?

Hi MadmanMurray!

I'm guessing that they want it in the form 6logx + 3logy.

(after all, how would you look up logx⁶ in log-tables if x = 2.345? … you'd have to find 2.345⁶ first, and the only way of doing that is to … yes! )

 

[MadmanMurray]

Thanks a lot.

I have 2 more log questions in front of me that are confusing as hell too:
1.) log (x² + 2) = 2.6

and

2.) 2^x + 1 = 3^{2x - 1}

For the first one there I was wondering if I can express it like this 2logx + 2? Can I do that?

 

[tiny-tim]

1.) log (x² + 2) = 2.6
…
I was wondering if I can express it like this 2logx + 2? Can I do that?

Nooo … that woud be (logx²) +2

Hint: if loga = b, then a = e^b

 

[jgens]

Careful tiny-tim: log(a) doesn't necessarily refer to the natural logarithm. log(a) commonly refers to the logarithm base 10 as well.

 

[Delphi51]

if log X = Y, then X = B^Y (where B is the base of the log)
If it doesn't make sense in logarithm land, transform it to power land!
And vice versa.

 

[kbaumen]

Careful tiny-tim: log(a) doesn't necessarily refer to the natural logarithm. log(a) commonly refers to the logarithm base 10 as well.

Well, where I study lg is decimal logarithm, ln is natural logarithm and log refers to a logarithm with any other base which is shown in subscript right after the log symbol. For example log$$_{2}$$8 = 3 ( I don't know why, but using LaTex here shows a subscript as a superscript on my machine. The 2 is supposed to be as a subscript. I hope you get the idea), lg100 = 2 and ln(e$$^{2}$$) = 2.

 

[tiny-tim]

For example log$_{2}$8 = 3 ( I don't know why, but using LaTex here shows a subscript as a superscript on my machine.

Hi kbaumen!

You have to use "inline" LaTeX (typing "itex" instead of "tex") if you're inserting into a line of text (see just above) …

but it's much better, on this forum, to use the X² or X₂ tags (just above the reply box), especially since any LaTeX takes up a lot of space on the server.

 

[kbaumen]

Hi kbaumen!

You have to use "inline" LaTeX (typing "itex" instead of "tex") if you're inserting into a line of text (see just above) …

but it's much better, on this forum, to use the X² or X₂ tags (just above the reply box), especially since any LaTeX takes up a lot of space on the server.

Oh. Thanks a lot for the explanation.

 

[HallsofIvy]

Thanks a lot.

I have 2 more log questions in front of me that are confusing as hell too:
1.) log (x² + 2) = 2.6

Then x²+ 2= a^2.6 where "a" is the base of the logarithm (probably 10 or e).

and

2.) 2^x + 1 = 3^{2x - 1}

Since there exponentials are to different bases, which cannot be converted to one another, there is no easy way to solve this equation.

For the first one there I was wondering if I can express it like this 2logx + 2? Can I do that?

 

[MadmanMurray]

Since there exponentials are to different bases, which cannot be converted to one another, there is no easy way to solve this equation.

I decided to plug in some random numbers and the first one I plugged in (1) happened to work. Since there's no simple way to solve it maybe that's what I was meant to do.