Express the given quantity as a single logarithm. 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2]

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Express the given quantity as a single logarithm. 1/5 ln(x + 2)^5 + 1/2 [ln x − ln(x^2 + 3x + 2)^2]

Mathematics
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\[\frac{ 1 }{ 5 }\log(x+2)^5+\frac{ 1 }{ 2 }[{\ln x- \ln (x^2+3x+2)^2}]= \frac{ 1 }{ 5 }*5 \log(x+2)+\frac{ 1 }{2 }[\ln x- \ln(x^2+3x+2)^2]\] i am sorry to say, this can't be unified because ln and log have different bases
All of them are base e, surely?
oops my bad

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Other answers:

Are there any nice identities for\[\log(a)^b\]? Maybe I'm being stupid.
Also- does anyone know how to have latex flow with your text rather than indent?
no you are right ...and it can be simplified
do you know the property of logarithm ? so like ln a + ln b = ? or ln a -ln b = ? or ln a^2 = ?
you dont like cooperating nothing ? i like help you here ... come on !!!
i'm confused
why ? so you know that ln a + ln b = ln a*b yes ?
yes
so than do you know how many will be ln a^2 =
or ln a - ln b =
ln(a/b)
yes and ln a^2 =
2ln a
yes right sure so than these property of logarithms can us in case of your exercise too ?
(ln(x + 2) (1/2 ln x / 2ln(x^2 + 3x + 2))
[ln(x + 2)] [1/2( ln x / 2ln(x^2 + 3x + 2))]
so the first part is right but the secondly you need to separate 1/5 ln(x+2)^5 = 5/5 ln(x+2) = ln(x+2) this is right so in the second part there are 1/2 (ln x - ln(x^2 +3x +2)^2) = 1/2 (ln (x/(x^2 +3x +2)^2) = ln (x/(x^2 +3x +2)^2)^(1/2) do you understand till now ? can you continue it ?
yess
so than how will be ?
[ln(x+2)] [ln (x/(x^2 +3x +2)^2)^(1/2)]
why ? than you know that (a^2)^(1/2) =a^(2/2) =a so than what will be ?
and you know that ln a +ln b = ln a*b
[ln(x+2)] [ln (x/(x^2 +3x +2)]
not is right please check it because there is just (x^2 +3x +2)^2 and x on numerator have not exponent 2 so what sign that not can be simplified by 1/2 OK ?
do you understand it sure ?
no
so the first part is right sure will be ln(x+2) OK . in the second part there are 1/2 (ln x - ln(x^2 +3x +2)^2) = 1/2 (ln (x/(x^2 +3x +2)^2)) = ln (x^(1/2) /((x^2 +3x +2)^2)^1/2 = ln (x^1/2) /(x^2 +3x +2) so can you continue it now ?
so than will result there ln (x+2) + ln (x^1/2) /(x^2 +3x +2) so from this can you continue it ?
are you here ?
so than will be using the property of logarithm log a +log b = log a*b so than will be ln ((x+2)*x^1/2 ) /(x^2 +3x +2)
so x^2 +3x +2 how can you factoriz it ?
do you know ?
so x^2 +3x +2 = (x+2)(x+1) is right ?
so than will be ln (x+2)x^(1/2) /(x+2)(x+1) so simplifie by (x+2) and will result ln x^1/2 /(x+1) so do you understand it now sure ?

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