displaystyleln243 Explanation: displaystyle1 . Begin by making use of the organic logarithmic property,displaystyleln_bleft(m^n ight)=ncdotln_bleft(m ight) ...

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depending on what you median by "exist", the answer come your concern is correct . There is an N=3 Poincaré supersymmetry algebra, and there room field-theoretic realisations. In particular there is a ...
as Claude Leibovici wrote, your factorization is false. Due to the fact that 3^x is fallout’s slowlier than 5^x we have actually to factor out 3^x and get: frac3^xln^2(3)+5^xln^2(5)3^xln(3)+5^xln(5)=frac3^xleft(ln^2(3)+left(frac53 ight)^xln^2(5) ight)3^xleft(ln(3)+left(frac53 ight)^xln(5) ight)=fracln^2(3)+left(frac53 ight)^xln^2(5)ln(3)+left(frac53 ight)^xln(5) ...
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together (a^b)^100 = a^100b because that all confident a and b, this difficulty is the exact same as notified 2^6, 3^5, 4^4, 5^3 and 6^2 i beg your pardon is right forward.
Area = int_1^3 left<(x+1) - frac2x ight>dx A = left|_1^3 If you advice , it gives 6-2ln(3) You have to calculate the area bounded through all 3 ...

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because that the second function, rewrite it as g(x) = -1 + frac21-x Then apply what you recognize for frac11-x, i beg your pardon is the geometric series -1 + 2cdotfrac11-x = -1 + 2sum_n=0^inftyx^n ...
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