Thread: Help With Log Notation

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

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

Thanks!

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

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

Originally Posted by mathguy20

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

$\displaystyle e \neq 10$ under any circumstance. In general however: $\displaystyle \ln(x) = \log_e(x)$ and $\displaystyle \log(x) = \log_{10}(x)$ where the base isn't explicitly stated

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

How would I solve for x?

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

We have: $\displaystyle Log_4{(x^2)}-Log_4{(x+3)}=1$

Can you see why?

So combine logs using the $\displaystyle Log(a)-Log(b)=Log(\frac{a}{b})$ and see if you get anywhere.

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

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 $\displaystyle 4 = \frac{x^2}{x+3}$. So $\displaystyle x = {-2, 6}$ ?

Originally Posted by mathguy20

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

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 $\displaystyle 4 = \frac{x^2}{x+3}$. So $\displaystyle x = {-2, 6}$ ?

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!

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.)

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.

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