Question

write the expression as a sum and or different logarithm. Express powers as factors, ln[x^2-x-20/(x+3)^3]^1/4

Answer 1

is it \[\ln[( x ^{2}-x-20) / (x+3)^{3}] ^{1/4}\] or \[\ln[( x ^{2}-x)-20 / (x+3)^{3}] ^{1/4}\]?
If it is the first case, then,
\[\ln[( x ^{2}-x-20) / (x+3)^{3}] ^{1/4}\]\[=1/4{ \ln[( x ^{2}-x-20) / (x+3)^{3}] }\]\[=1/4 [\ln( x ^{2}-x-20)- \ln(x+3)^{3}] \]\[=1/4 \ln( x ^{2}-x-20)- 3/4 \ln(x+3) \]

Answer 2

so whts the ans?

Answer 3

which was the case then? ln[(x^2 −x−20)/(x+3) 3 ]^( 1/4 ) or ln[(x^ 2 −x)−20/(x+3) 3 ] ^(1/4 ) ?

Answer 4

the first case

Answer 5

=1/4ln(x^2 −x−20)−3/4ln(x+3)

Answer 6

use log rules
log(x^n) = nlog(x)
log(x)+log(y) = log(xy)
log(x) -log(y) = log(x/y)

Answer 7

for this question i guess using
log(x^n) = nlog(x) and log(x) -log(y) = log(x/y) is enough to get the answer

Answer 8

factor x2-x-20
=(x-5)(x+4)

Answer 9

i need an answer pls

Answer 10

huh , i see, so =1/4ln(x-5)+1/4ln(x+4)-3/4ln(x+3) , is that right?

Answer 11

yes

Answer 12

thank you so much!!!

Answer 13

no, thnk you lol

Answer 14

no, i did learn a lot from this :)