Help With Log Notation

• May 2nd 2011, 04:05 PM
mathguy20
Help With Log Notation
What would $ln$ stand for in a log equation?

For example $8ln x = 16$ ? Is that the same as $8 log x = 16$ ?

Thanks!
• May 2nd 2011, 04:06 PM
e^(i*pi)
$\ln(x) = \log_e(x)$
• May 2nd 2011, 04:12 PM
mathguy20
Thanks for the quick response! Does $e$ = 10 is the base if it not stated, just like normal logs?
• May 2nd 2011, 04:17 PM
e^(i*pi)
Quote:

Originally Posted by mathguy20
Thanks for the quick response! Does $e$ = 10 is the base if it not stated, just like normal logs?

$e \neq 10$ under any circumstance. In general however: $\ln(x) = \log_e(x)$ and $\log(x) = \log_{10}(x)$ where the base isn't explicitly stated
• May 2nd 2011, 04:20 PM
mathguy20
Got it -- here's another log question if you don't mind.

How would I solve for x?

$2log_4(x) - log_4(x+3) = 1$
• May 2nd 2011, 04:27 PM
Quacky
We have: $Log_4{(x^2)}-Log_4{(x+3)}=1$

Can you see why?

So combine logs using the $Log(a)-Log(b)=Log(\frac{a}{b})$ and see if you get anywhere.
• May 2nd 2011, 04:36 PM
mathguy20
Yeah - so would it be $Log_4(\frac {x^2}{x+3}) = 1$ ?

I think I got lost in the base 4, because I've only done those in base 10.

Then, I can just solve for x from $4 = \frac{x^2}{x+3}$. So $x = {-2, 6}$ ?
• May 2nd 2011, 04:41 PM
Quacky
Quote:

Originally Posted by mathguy20
Yeah - so would it be $Log_4(\frac {x^2}{x+3}) = 1$ ?

I think I got lost in the base 4, because I've only done those in base 10.

Then, I can just solve for x from $4 = \frac{x^2}{x+3}$. So $x = {-2, 6}$ ?

Yeah. Remember that you can always try it and substitute your answer into the original equation to check!

Edit: although x can't be -2 because this isn't defined by your original equation. Be careful with logs; this happens frequently!
• May 2nd 2011, 04:43 PM
mathguy20
Quote:

Originally Posted by Quacky
Yeah. Remember that you can always try it and substitute your answer into the original equation to check!

I see how I can plug in -2 and 6 to the equation we got in the final steps, but is there a way I can plug that into the original equation and check it on my TI-84? (I'm not sure how to use bases other than 10 on the calculator.)
• May 2nd 2011, 04:48 PM
Quacky
I am unfamiliar with this exact calculator, but I would expect, if there isn't a button that calculates logs to all bases, that you can, using the change of base rule. If you convert the logs from base 4 to base 10, then you can plug the numbers in. I don't know whether there's a shorthand, my calculator can calculate logs to all bases for me so I don't have to worry.(Cool)
• May 2nd 2011, 07:12 PM
bjhopper
Hello mathguy20,
log B4 (4) =1. Bring all terms to LHS.Combine all the log terms to 1 term= to 0.Solve for x

bjh