Logarithm equations home

In summary, the conversation discusses three equations that the individual is having difficulty solving. The equations involve logarithms and substitutions to solve for the variable. The individual also clarifies that "lg" represents the logarithm with a base of 10. They also mention that the solutions to two of the equations are x = -1 and x = 1000 or x = 100. However, the individual clarifies that the type of logarithm should be specified as there are different bases that can be used.
  • #1
Kjos
5
0

Homework Statement



i.2(lgx)^2 - lgx = 0

and

ii. lg(2x-2)^2= 4lg(1-x)

and

iii. lgx-6 / lgx-4 = lgx.

I simply do not manage to solve these equations, and I would therefore be happy for all help. Thanks in advance.
 
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  • #2
Hints:

i) and iii) Make the substitution z = lg x, and solve for z. What do you get?

ii) Notice that: lg (2x - 2) = lg[(-2)(1 - x)]. Now, if lg represents a principal value of the logarithm with a branch cut along the negative real axis, then:

[tex]
\mathrm{lg} \left[(-2)(1 - x) \right] = \mathrm{lg}(e) \cdot \ln{\left[(-2)(1-x) \right]} = \left\lbrace
\begin{array}{ll}
\mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x)] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) &, \ \mathrm{Arg}(1-x) \le 0 \\

\mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x) - 2 \pi \, i] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) - 2 \pi \, \mathrm{lg}(e) \, i &, \ \mathrm{Arg}(1-x) > 0
\end{array}\right.
[/tex]

Then, substitute z = lg(1 - x) and solve the equsation for z. Make sure that the condition for Arg(1 - x) is satisfied in the proper case.
 
  • #3
Dickfore said:
Hints:

ii) Notice that: lg (2x - 2) = lg[(-2)(1 - x)]. Now, if lg represents a principal value of the logarithm with a branch cut along the negative real axis, then:

[tex]
\mathrm{lg} \left[(-2)(1 - x) \right] = \mathrm{lg}(e) \cdot \ln{\left[(-2)(1-x) \right]} = \left\lbrace
\begin{array}{ll}
\mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x)] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) &, \ \mathrm{Arg}(1-x) \le 0 \\

\mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x) - 2 \pi \, i] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) - 2 \pi \, \mathrm{lg}(e) \, i &, \ \mathrm{Arg}(1-x) > 0
\end{array}\right.
[/tex]

This is the only one I can't figure out. Isn't there an easier way?
 
  • #4
First of all, what do you mean by "lg"? Second, what is the domain in which you solve this equations? Is it complex numbers, or are you restricted to real numbers only?
 
  • #5
Dickfore said:
First of all, what do you mean by "lg"? Second, what is the domain in which you solve this equations? Is it complex numbers, or are you restricted to real numbers only?


Lg = Logarithm
Real numbers only, but no worries, I think I worked it out.

Thanks!
 
  • #6
Kjos said:

Homework Statement



i.2(lgx)^2 - lgx = 0

and

ii. lg(2x-2)^2= 4lg(1-x)

and

iii. lgx-6 / lgx-4 = lgx.

I simply do not manage to solve these equations, and I would therefore be happy for all help. Thanks in advance.

(ii) Is ambiguous. Do you mean
[lg(2x-2)]^2 = 4 lg(1-x), or do you mean
lg[(x-2)^2] = 4 lg(1-x)?

(iii) As written, (iii) is
(lg x) - (6/lg x) - 4 = lg x. Did you mean that? Or, did you mean
lg(x-1)/lg(x-4) = lg x? Or did you mean
[lg(x-1)/lg x] - 4 = lg x?

You need to use brackets to make things clear.

RGV
 
  • #7
Ray Vickson said:
(ii) Is ambiguous. Do you mean
[lg(2x-2)]^2 = 4 lg(1-x), or do you mean
lg[(x-2)^2] = 4 lg(1-x)?

(iii) As written, (iii) is
(lg x) - (6/lg x) - 4 = lg x. Did you mean that? Or, did you mean
lg(x-1)/lg(x-4) = lg x? Or did you mean
[lg(x-1)/lg x] - 4 = lg x?

You need to use brackets to make things clear.

RGV

My mistake. Check the link. http//bildr.no/view/1297631
 
Last edited:
  • #8
Kjos said:
My mistake. [PLAIN]http://bildr.no/view/1297631[/QUOTE] [Broken]

So, what are the answers to the questions I asked? Also, someone asked you what "lg" represents, and you said "logarithm". But, there are three kinds of logarithms most commonly used : (i) logarithms to base 2; (ii) logarithms to base 10; and (ii) natural logarithms. Just saying "logarithm" does not really help us; you need to specify what type of logarithm you mean.

RGV
 
Last edited by a moderator:
  • #9
Ray Vickson said:
So, what are the answers to the questions I asked? Also, someone asked you what "lg" represents, and you said "logarithm". But, there are three kinds of logarithms most commonly used : (i) logarithms to base 2; (ii) logarithms to base 10; and (ii) natural logarithms. Just saying "logarithm" does not really help us; you need to specify what type of logarithm you mean.

RGV

I don't know actually, but I think it is logarithms to base 10.

And regarding you question, please check out the link attached above.

The correct answers are:

iii. X = 1000 or X = 100

ii. x =-1

However, it is not that important since I've got until monday.
 
  • #10
If the symbol is given as logx it is the default one, with the base of 10. In other cases you have the logarithm given as lognX - where n is the base and it is equal to log10X/log10n and finally you have the natural logarithm ln X which has a base of e - so ln X = logeX. Hope that clarifies it for you :)
 
  • #11
lendav_rott said:
If the symbol is given as logx it is the default one, with the base of 10. In other cases you have the logarithm given as lognX - where n is the base and it is equal to log10X/log10n and finally you have the natural logarithm ln X which has a base of e - so ln X = logeX. Hope that clarifies it for you :)

Unfortunately, what you say is not completely correct: many authors use log x to mean ln x, etc., so you really do need to check. Often if one wants to use a base other than e, one uses a subscript like log10(x) or log2(x) or an index notation like log[10](x), etc. I think usage has changed a lot during the more than 50 years since I was in school; then log(x) really did always mean log[10](x). Now, not so much.

RGV
 

1. What is a logarithm equation?

A logarithm equation is a mathematical expression that uses logarithms, which are the inverse functions of exponential functions. It is used to solve for the unknown variable in an exponential equation.

2. What is the purpose of logarithm equations in a home setting?

Logarithm equations are commonly used in home settings for financial planning, such as calculating interest rates, investments, and mortgage payments. They can also be used to solve for time and growth rates in home improvement projects.

3. How do I solve a logarithm equation at home?

To solve a logarithm equation at home, you can use a calculator or follow a step-by-step process. First, isolate the logarithm term on one side of the equation. Then, convert the logarithm into an exponential form. Finally, solve for the unknown variable using basic algebraic principles.

4. Are there any common mistakes to avoid when solving logarithm equations at home?

Yes, some common mistakes when solving logarithm equations at home include forgetting to convert the logarithm into exponential form, incorrectly using the rules of logarithms, and not checking for extraneous solutions. It is important to double-check your work and use a calculator if needed.

5. What are some real-life applications of logarithm equations in a home setting?

Aside from financial planning and home improvement projects, logarithm equations can also be used to calculate pH levels in pools and hot tubs, measure decibels in home sound systems, and determine the half-life of household cleaning products.

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