Solving log₂(x) + log₄(5) = -1 without a calculator

[pat666]

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

 

[rock.freak667]

Homework Equations

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

Also log_ax=log_ay ⇒ x=y

 

[pat666]

ok ill have a look into that and get back to you if i stil can't do it... thanks

 

[Mentallic]

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 $$2^{-1}=a$$

 

[pat666]

so it will go to log5/log4 base 2 obviously and then what do i do??

 

[Mentallic]

Yep, can that be simplified at all? What is log₂4?

 

[pat666]

=log_2 2^2 = 2??

 

[Mentallic]

Correct, so now we have $$log_2x+\frac{log_25}{2}=-1$$

Now use the rule that $$a.log_bc=log_b(c^a)$$

 

[pat666]

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?

 

[Mentallic]

No no, use it on that one term $$\frac{log_25}{2}$$. Think about what the 'a' in $$alog_bc$$ is here.

 

[pat666]

i think the a is 1/2. from that =log_2 (5^.5) still not sure how to proceed from here though?

 

[Mentallic]

Good, now look at what you have:

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

The left part be simplified into one log

 

[pat666]

ok i got it from here thanks for that.

 

[Mentallic]

You're welcome.