v.s
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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ghazi
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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
henpen
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All of them are base e, surely?
ghazi
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oops my bad
henpen
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Are there any nice identities for\[\log(a)^b\]? Maybe I'm being stupid.
henpen
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Also- does anyone know how to have latex flow with your text rather than indent?
ghazi
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no you are right ...and it can be simplified
jhonyy9
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do you know the property of logarithm ?
so like ln a + ln b = ?
or ln a -ln b = ?
or ln a^2 = ?
jhonyy9
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you dont like cooperating nothing ?
i like help you here ... come on !!!
v.s
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i'm confused
jhonyy9
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why ? so you know that ln a + ln b = ln a*b
yes ?
v.s
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yes
jhonyy9
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so than do you know how many will be ln a^2 =
jhonyy9
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or ln a - ln b =
v.s
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ln(a/b)
jhonyy9
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yes and ln a^2 =
v.s
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2ln a
jhonyy9
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yes right sure
so than these property of logarithms can us in case of your exercise too ?
v.s
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(ln(x + 2) (1/2 ln x / 2ln(x^2 + 3x + 2))
v.s
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[ln(x + 2)] [1/2( ln x / 2ln(x^2 + 3x + 2))]
jhonyy9
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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 ?
v.s
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yess
jhonyy9
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so than how will be ?
v.s
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[ln(x+2)] [ln (x/(x^2 +3x +2)^2)^(1/2)]
jhonyy9
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why ? than you know that (a^2)^(1/2) =a^(2/2) =a
so than what will be ?
jhonyy9
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and you know that ln a +ln b = ln a*b
v.s
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[ln(x+2)] [ln (x/(x^2 +3x +2)]
jhonyy9
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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 ?
jhonyy9
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do you understand it sure ?
v.s
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no
jhonyy9
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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 ?
jhonyy9
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so than will result there ln (x+2) + ln (x^1/2) /(x^2 +3x +2)
so from this can you continue it ?
jhonyy9
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are you here ?
jhonyy9
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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)
jhonyy9
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so x^2 +3x +2 how can you factoriz it ?
jhonyy9
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do you know ?
jhonyy9
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so x^2 +3x +2 = (x+2)(x+1)
is right ?
jhonyy9
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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 ?