Please help show how to solve..

Question

anonymous · Answer

$$\log_{5} 3$$

anonymous · Answer

$$\log_{5} 4$$

anonymous · Answer

@jim_thompson5910

e.mccormick · Answer

You know what the relationship between the base of 5 and the 4 or 3 is?

anonymous · Answer

no

anonymous · Answer

im completely lost

e.mccormick · Answer

Ah.  OK.  

If you look at $y = log_b(x)$ it means b to some power = x.

e.mccormick · Answer

The log solves for what that power is.

anonymous · Answer

Ok
so where does the log comes in?

e.mccormick · Answer

Lets start with a little easuer one then apply it to what you have here.  Say: $$y=\log_{10}(1000)$$
That means 10 to some power=1000.  That some power is y.  So another way if writing the same thing is this:$$10^y=1000$$

anonymous · Answer

y=3?

e.mccormick · Answer

In that case, yes. Because it is an easy one.  Now, lets look at the harder one you have.
$$y=\log_5(4)$$There is no easy way to just say, Oh!  $5^y=4$  means y is...
The solution to that problem is the change of base formula.  Have you studied that?

anonymous · Answer

havent studied it yet

e.mccormick · Answer

OK.  Let me go over that.  On your calculator there are two and only two logs, $\log_e$ and $\log_{10}$.  If it is not in either of those, you can not get a numeric approximation for the answer!  That is where the change of base comes in.

e.mccormick · Answer

$$\log_b(x)=\frac{\log_n(x)}{\log_n(b)}$$Where $n=\{e,10\}$
That lets you recast the question into something you can solve with a calculator.

anonymous · Answer

Ok thanks

e.mccormick · Answer

So, what did you get for an answer?

anonymous · Answer

.86?

e.mccormick · Answer

Yep!  Looks like you understood just fine.  Have fun!

anonymous · Answer

ok how do i do $$\log_{5} (4) + \log_{5} (3)$$

anonymous · Answer

solve for the first and then the second then add?

e.mccormick · Answer

AH!  Now there you have the other rules of logs.  That one you might even be able to get a more accurate answer for.

e.mccormick · Answer

Here are the three basic rules of logs: http://www.purplemath.com/modules/logrules.htm If you look at rule #1, you can see that you can multiply 4 and 3 to have just one log to work with, which reduces the number of errors you could make. It also means any calculator approximation is done last, so there is less error there too.

anonymous · Answer

OK thanks for your help. ill go read up on it

e.mccormick · Answer

Yah, I like purple math as a complement to the book.  But you can see pretty easy that what you have with $\log_5(4)+\log_5(3)=\log_5(12)$  That is less work in the calculator.

e.mccormick · Answer

One word of waring:  The bases MUST match for you to apply these rules!