1. ## Natural logarithm problem

I am studying for a math competition at my school and this here just stumps me.
$\displaystyle \textrm{ln}\frac{3x^4}{5y^2}$

My work:
Quotient property:

$\displaystyle \textrm{ln}3x^4-\textrm{ln}5y^2$

Power property:

$\displaystyle 4\textrm{ln}3x-2\textrm{ln}5y$

Product property:

$\displaystyle \textrm{ln}3+4\textrm{ln}x-\textrm{ln}5+2\textrm{ln}y$

And this is what I have got after a long time of trying...

$\displaystyle \textrm{ln}3+4\textrm{ln}x-\textrm{ln}5-2\textrm{ln}y$
can someone explain this to me?
thanks.

$\displaystyle \int$

2. $\displaystyle \text{ln}ab-\text{ln}cd= \text{ln}a+\text{ln}b-(\text{ln}c+\text{ln}d)$, not $\displaystyle \text{ln}a+\text{ln}b-\text{ln}c+\text{ln}d$

3. Originally Posted by integral
$\displaystyle \textrm{ln}\frac{3x^4}{5y^2}$

My work:
Quotient property:

$\displaystyle \textrm{ln}3x^4-\textrm{ln}5y^2$

Power property:

$\displaystyle 4\textrm{ln}3x-2\textrm{ln}5y$
Not sure if this is correct,

Is is true that $\displaystyle \ln(a)^b = b\ln(a)$

But you have $\displaystyle \ln(3x^4) = \ln(3)+\ln(x^4) = \ln(3)+4\ln(x)$

You have somehow ended up with the correct working but only by fluke.

What is the question actually asking?

4. Originally Posted by pickslides
What is the question actually asking?
I think it was asking to express $\displaystyle \textrm{ln}\frac{3x^4}{5y^2}$ in terms of $\displaystyle x$ and $\displaystyle y$. So we have: $\displaystyle \textrm{ln}\frac{3x^4}{5y^2} = \textrm{ln}{3x^4}-\textrm{ln}{5y^2} = \textrm{ln}{3}+\textrm{ln}{x^4}-(\textrm{ln}{5}+\textrm{ln}{y^2}) = \textrm{ln}{3}+\textrm{4ln}{x}-(\textrm{ln}{5}+\textrm{2ln}{y})$, but the OP thought it as $\displaystyle \textrm{ln}{3}+\textrm{4ln}{x}-\textrm{ln}{5}+\textrm{2ln}{y}$ instead of $\displaystyle \textrm{ln}{3}+\textrm{4ln}{x}-\textrm{ln}{5}-\textrm{2ln}{y}.$

5. Originally Posted by pickslides
Not sure if this is correct,

Is is true that $\displaystyle \ln(a)^b = b\ln(a)$

But you have $\displaystyle \ln(3x^4) = \ln(3)+\ln(x^4) = \ln(3)+4\ln(x)$
What? product rule states: $\displaystyle log_buv=log_bu+log_bv$ So if I did: $\displaystyle ln(3x^4)=4ln(3x)$ it would still end up as
$\displaystyle \ln(3)+4\ln(x)$
You have somehow ended up with the correct working but only by fluke.

What is the question actually asking?
.

6. I think someone above has already corrected the error. I personally think that you made a tiny mistake after the above stage. It should be
$\displaystyle =\textrm{ln}3+4\textrm{ln}x-(\textrm{ln}5 + \textrm{ln}y^2)$
The negative applies to both of the two log functions at the end, just as it does at the beginning of my post when they are combined.
This then becomes the correct answer of: