How Do You Solve These Logarithm Equations?

[Kjos]

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.

 

[Dickfore]

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:

<br /> \mathrm{lg} \left[(-2)(1 - x) \right] = \mathrm{lg}(e) \cdot \ln{\left[(-2)(1-x) \right]} = \left\lbrace<br /> \begin{array}{ll}<br /> \mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x)] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) &amp;, \ \mathrm{Arg}(1-x) \le 0 \\<br /> <br /> \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 &amp;, \ \mathrm{Arg}(1-x) &gt; 0<br /> \end{array}\right.<br />

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.

 

[Kjos]

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:

<br /> \mathrm{lg} \left[(-2)(1 - x) \right] = \mathrm{lg}(e) \cdot \ln{\left[(-2)(1-x) \right]} = \left\lbrace<br /> \begin{array}{ll}<br /> \mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x)] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) &amp;, \ \mathrm{Arg}(1-x) \le 0 \\<br /> <br /> \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 &amp;, \ \mathrm{Arg}(1-x) &gt; 0<br /> \end{array}\right.<br />

This is the only one I can't figure out. Isn't there an easier way?

 

[Dickfore]

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?

 

[Kjos]

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!

 

[Ray Vickson]

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

 

[Kjos]

(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

 

[Ray Vickson]

My mistake. [PLAIN]http://bildr.no/view/1297631[/QUOTE]

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

 

[Kjos]

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.

 

[lendav_rott]

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 log_nX - 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 = log_eX. Hope that clarifies it for you :)

 

[Ray Vickson]

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 log_nX - 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 = log_eX. 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