# Logs no caculator

1. Jun 4, 2010

### pat666

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

2. Jun 4, 2010

### rock.freak667

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

3. Jun 4, 2010

### pat666

ok ill have a look into that and get back to you if i stil cant do it... thanks

4. Jun 4, 2010

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

5. Jun 4, 2010

### pat666

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

6. Jun 5, 2010

### Mentallic

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

7. Jun 5, 2010

### pat666

=log_2 2^2 = 2??

8. Jun 5, 2010

### Mentallic

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

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

9. Jun 5, 2010

### pat666

ok im stumped here i cant get it into a form where that can be used. log_2x=-1-log_2 5/2???

10. Jun 5, 2010

### Mentallic

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

11. Jun 5, 2010

### pat666

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

12. Jun 5, 2010

### Mentallic

Good, now look at what you have:

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

The left part be simplified into one log

13. Jun 5, 2010

### pat666

ok i got it from here thanks for that.

14. Jun 5, 2010

### Mentallic

You're welcome.

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