How to Solve These Logarithm Equations?

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Homework Help Overview

The discussion revolves around solving logarithmic equations, specifically focusing on two problems involving logarithmic identities and properties. The original poster expresses uncertainty in their approach to these equations, which are part of an assignment not covered in class.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate logarithmic expressions but encounters confusion regarding the validity of their steps. They express uncertainty about how to proceed after reaching certain points in their calculations.
  • Some participants question the legality of the operations performed by the original poster, particularly in the transition from one line of their work to another.
  • Others suggest using the relationship between logarithmic and exponential functions to solve for the variable x, prompting further inquiry into the original poster's understanding of these concepts.
  • There is a discussion about the implications of the original poster's equations, with one participant noting a misunderstanding in the interpretation of an equation as an identity.

Discussion Status

The discussion is ongoing, with participants providing clarifications and hints without offering direct solutions. The original poster has acknowledged their confusion and is actively seeking guidance on how to resolve the issues raised in their attempts.

Contextual Notes

The original poster mentions that the questions were given as an assignment during a break, indicating a potential lack of prior instruction on the relevant material. This context may contribute to the uncertainties expressed in the discussion.

Nx2
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hi guys, I am not too sure how to do these questions:
solve for x,
a) (logx^3)^2 = logx^18
b) logx^3 + log(x^logx) = -2

so this is what i got so far:
a) (3logx)^2 = logx^18
9logx^2 = logx^18
18logx = logx^18
logx^18 = logx^18
... then i got stuck... i was clueless and did not know how to solve for x.

b) 3logx + (logx)(logx) = -2
logx^2 + 3logx +2 = 0
so let logx = s
s^2 +3s + 2 = 0
(s+1)(s+2)=0
s=-1 or s=-2
... then again i got stuck... I am not sure if what i did was right. our teacher gave us these questions that she has never taught us b4 and said this is ur assignment for the break... any help would be very much appreciated. thnx.

- Tu
 
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a) You did an illegal operation btw line 2 and line 3 when you said 9logx^2 = 18logx

What you got here is [itex]9(logx)^2[/itex] and not [itex]9log(x^2)[/itex]. And while it is true that [itex]9log(x^2)=18logx[/itex], it is not that [itex]9(logx)^2=18logx[/itex]
 
b) You're almost there! you got the solutions s=-1 and s=-2, which translate into logx =-1 and logx=-2. Now you got to solve for x. Hint: Use the fact that [itex]e^x[/itex] is the inverse function of [itex]logx[/itex]. This means that

[tex]e^{logx}=x[/tex]

mmh. :smile:
 
oo... ok... so now that i have that... how do i solve for x in a though?
 
Nx2 said:
oo... ok... so now that i have that... how do i solve for x though?
see my last post. same trick.
 
srry.. but i don't think we learned e^x
 
Mmmh, did you learn about exp(x) ?
 
Last edited:
yea i know exp(x)
 
They are the same.

[tex]\mbox{exp}(x) = e^x[/tex]

So you know that exp(x) is the inverse function of logx, right?

Then exp(logx)=x.
 
  • #10
ok.. but i still don't get how i would solve for x if i had
logx^18 = logx^18... wouldn't i just end up with 0 = 0?
 
  • #11
err. you don't have logx^18 = logx^18. I told you there was a mistake between line 2 and 3. All you have is

[tex]9(logx)^2 = (logx)^{18}[/tex]

If you don't see where to go from there, use the same substitution as you did in b). Set s = logx... then solve for s. Then use the fact that exp(s)=x to solve for x.
 
Last edited:
  • #12
omg... that's why it wasnt working... i was like so clueless... forgot bout that... thnx a lot man. i c what u meant when u were talking bout e^x now... thnx, very much appreciated for putting up with me.
 
  • #13
It's all good. Don't hesistate to post if something doesn't work out!
 

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