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

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To solve the equation log_2(x) + log_4(5) = -1 without a calculator, it's essential to apply the product law for logarithms and the change of base formula. Transform log_4(5) into a base 2 logarithm, leading to log_2(5)/2. This allows the equation to be simplified to log_2(x) + log_2(√5) = -1. The left side can then be combined into a single logarithm, enabling the final steps to isolate x. The discussion emphasizes the importance of understanding logarithmic properties to solve the equation effectively.
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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,

log_ab=\frac{log_cb}{log_ca} 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 log_2a=-1, so 2^{-1}=a
 
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 log_2x+\frac{log_25}{2}=-1

Now use the rule that a.log_bc=log_b(c^a)
 
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 \frac{log_25}{2}. Think about what the 'a' in alog_bc 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:

log_2x+log_2(\sqrt{5})=-1

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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