Help With Log Notation

**mathguy20** · May 2nd 2011, 05:05 PM

What would $ln$ stand for in a log equation?

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

Thanks!

**e^(i*pi)** · May 2nd 2011, 05:06 PM

$\ln(x) = \log_e(x)$

**mathguy20** · May 2nd 2011, 05:12 PM

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

**e^(i*pi)** · May 2nd 2011, 05:17 PM

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

**mathguy20** · May 2nd 2011, 05:20 PM

Got it -- here's another log question if you don't mind.

How would I solve for x?

$2log_4(x) - log_4(x+3) = 1$

**Quacky** · May 2nd 2011, 05:27 PM

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.

**mathguy20** · May 2nd 2011, 05:36 PM

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 $x = {-2, 6}$ ?

**Quacky** · May 2nd 2011, 05:41 PM

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!

**mathguy20** · May 2nd 2011, 05:43 PM

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.)

**Quacky** · May 2nd 2011, 05:48 PM

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.

**bjhopper** · May 2nd 2011, 08:12 PM

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

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