Need Help Please! I know what I'm supposed to do but keep getting the wrong answer =/

Use the laws of logarithms to expand and simplify the expression.
ln((x^2)/(square root x (1+4)^4))

Question

Need Help Please! I know what I'm supposed to do but keep getting the wrong answer =/

Use the laws of logarithms to expand and simplify the expression. 
ln((x^2)/(square root x (1+4)^4))

vane11 · Answer

$$\ln ((x ^{2})/(\sqrt{x} (1+x)^{4})$$
I messed up the first time I wrote the equation this is the correct one

ganeshie8 · Answer

$$\large \ln \left(\dfrac{x^2}{\sqrt{x}(1+x)^4}\right)$$

ganeshie8 · Answer

like that ?

vane11 · Answer

Yeah I tried making it look nicer haha

ganeshie8 · Answer

good, do u knw any log rules ?

ganeshie8 · Answer

http://www.chilimath.com/algebra/advanced/log/images/466x428xrules,P20of,P20exponents.gif.pagespeed.ic.1z7-Ekl5L5.png

vishweshshrimali5 · Answer

@Vane11, Ganeshi means these:

(1)$\large{\ln(\cfrac{a}{b}) = \ln{a} - \ln{b}}$

(2) $\large{\ln(ab) = \ln {a} + \ln{b}}$

(3) $\large{\ln(a^b) = b\ln a}$

vane11 · Answer

Yes I do :)

vishweshshrimali5 · Answer

Good:

then apply them in the given order.

vane11 · Answer

the website is acting strange takes forever and freezes sorry for the delays btw

vishweshshrimali5 · Answer

Yeah. It happens even with me. :)

vishweshshrimali5 · Answer

So, first use the law 1 to change the division into subtraction of 2 logarithms.

vishweshshrimali5 · Answer

For example:

See:

$$\large \ln \left(\dfrac{x^2}{\sqrt{x}(1+x)^4}\right)$$

$$\large{ = \ln (x^2) - \ln(\sqrt{x}(1+x)^4)}$$

vane11 · Answer

so then 2lnx for the first part right?

vishweshshrimali5 · Answer

yes

vane11 · Answer

I know sqrt is exponent 1/2 but idk if im supposed to separate the second half into two sections: -1/2lnx - 4ln(1+x) I feel that's wrong though

vishweshshrimali5 · Answer

well I don't think there is anything wrong in it

vane11 · Answer

Yeah I put that answer down and its marked wrong, I've tried playing with the parenthesis and leaving the second half as is as well as the first but no combination has worked for me

vishweshshrimali5 · Answer

$$\large{2\ln{x} - \cfrac{1}{2}\ln{x} - 4\ln(1+x)}$$

$$\large{= \cfrac{3}{2}\ln{x} - 4\ln(1+x)}$$

vane11 · Answer

hm good point like terms hadn't thought of that

vishweshshrimali5 · Answer

try this

vane11 · Answer

it worked! I guess it was the expand AND simplify that I was leaving out, I'd expand it but forgot about combining like terms, thanks! definitely a good lesson to remember for the exam :)

vishweshshrimali5 · Answer

:)