Solve Logs without a Calculator: log_2(x)+log_4(5)=-1

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Homework Help Overview

The discussion revolves around solving a logarithmic equation involving different bases, specifically log_2(x) + log_4(5) = -1, without the use of calculators. The original poster expresses difficulty due to a lack of recent experience with logarithms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of logarithmic properties, such as the product law and change of base rule. There are attempts to rewrite log_4(5) in terms of log_2 and to simplify the equation. Questions arise regarding the simplification of logarithmic expressions and the manipulation of terms.

Discussion Status

Participants are actively engaging with the problem, offering hints and guidance on how to approach the equation. There is a progression in the discussion, with some participants successfully simplifying parts of the equation, while others express uncertainty about the next steps. No explicit consensus has been reached, but productive dialogue is ongoing.

Contextual Notes

The original poster is constrained by the requirement to solve the problem without a calculator, which adds to the challenge of recalling logarithmic properties and manipulation techniques.

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Homework Statement


log_2(x)+log_4(5)=-1 just found out we have to do logs with no calculators and its been about 2 years since i last had anything to do with them. anyway i need some help to solve this.


Homework Equations





The Attempt at a Solution

 
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pat666 said:

Homework Equations


You will need to find these. Mainly look into the product law for logs and the change of base rule.

Also logax=logay ⇒ x=y
 
ok ill have a look into that and get back to you if i stil can't do it... thanks
 
The rule you need for this question are the change of base rules,

[tex]log_ab=\frac{log_cb}{log_ca}[/tex] where c>0 and any base you choose it to be.

Now change your log45 into a log base 2, and from there you can use your other rules of logarithms to turn the equation into [tex]log_2a=-1[/tex], so [tex]2^{-1}=a[/tex]
 
so it will go to log5/log4 base 2 obviously and then what do i do??
 
Yep, can that be simplified at all? What is log24?
 
=log_2 2^2 = 2??
 
Correct, so now we have [tex]log_2x+\frac{log_25}{2}=-1[/tex]

Now use the rule that [tex]a.log_bc=log_b(c^a)[/tex]
 
ok I am stumped here i can't get it into a form where that can be used. log_2x=-1-log_2 5/2?
 
  • #10
No no, use it on that one term [tex]\frac{log_25}{2}[/tex]. Think about what the 'a' in [tex]alog_bc[/tex] is here.
 
  • #11
i think the a is 1/2. from that =log_2 (5^.5) still not sure how to proceed from here though?
 
  • #12
Good, now look at what you have:

[tex]log_2x+log_2(\sqrt{5})=-1[/tex]

The left part be simplified into one log :smile:
 
  • #13
ok i got it from here thanks for that.
 
  • #14
You're welcome.
 

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