# Homework Help: Confused about basic log formula

1. Feb 11, 2017

### late347

1. The problem statement, all variables and given/known data
$$ln(x^2)=4$$

2. Relevant equations
$a^{log_a(x)}=x$
$log_a(a^x)=x$

3. The attempt at a solution

ln(x^2)=4

<=> $2ln(x)=ln(4)$

<=> $ln(x)= [ln(4)]/2$

<=> $log_e(x)= [ln(4)]/2$

<=> $e^{ln(4)/2}=x$

<=> $[e^{ln(4)}]^{1/2}$

<=> $sqrt(e^{ln(4)})$

here I was a little bit confused about, how can we know what the thing inside the square root will be, in order to take the square root from it?
What do we know about the value of (e^{ln(4)}) so the square root of it can be taken?

I understand the other formula which was
$log_a(a^x)=x$
obviously the $log_a(a^x)$ asks us what exponent is the correct one, when you want to raise a to the exponent of something, such that the result will become a^x. The answer is x for the exponent.

2. Feb 11, 2017

### late347

oops I made a terrible mistake there
It helps to read with your brain and eyes. Instead of just eyes and not using the brain.

ln(x^2)=4

<==>
log_e(x^2) =4 {at this point I think we must keep the squared term instead of using dropping the exponent down such as log(x^r) = r*log(x)}
<==>
e^4= x^2
<==>
x= ± (e^2)

3. Feb 11, 2017

### Ray Vickson

Alternatively, you could write $|x| = e^2.$

4. Feb 12, 2017

### late347

What about the log formula being used at another example such as

$e^{log_e(4)÷2}=???$

I have a harder time justifying that... I know that the result is 2 because I used a calculator.

You could arrange that such as
$sqrt(e^{ln(4)})$

Hmmm... now that I think about it more...

It seems that ln4 means... the exponent (lets call it with variable r ) with the requirement that it is the specific exponent which is required to be used to raise (e^1) such that the result of e^r=4

So therefore it seems that the intermediate result in the exponent spot there... e^{ln4} = 4
Because e^r=4.

And square root of 4 is 2