I need someone who knows there Math:  How do I simplify:  ln e ^m ?
951745
ln e^m = m\n\ne^(ln x^4) = x^4 as e^(in z) = z\n\nlog (base 2) 8 = 3 as 2^3 =8
1132135
I think you need to rephrase this question.  I'm going to solve this problem:\n\n6^(2x)>5^(x+2)\n\nFirst, logarithm both sides and simplify.\n\n2x*ln6>(x+2)*ln5\n\nln5 and ln6 are both positive, so you don't have to worry about changing your sign anytime in the future.  Now just get all of the x's on one side and factor.\n\nx*(2ln6-ln5)>2ln5\n\n2ln6-ln5>0, so no worries in solving for x.\n\nx>(2ln5)/(2ln6-ln5)\n\nor \n\nx>1.631
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
458822
There's three log rules you need to know:\n\n1) ln a + ln b = ln (ab)\n\n2) ln a - ln b = ln (a/b)\n\n3) ln a^b = b×ln a\n   where "^" means "to the power of"\n\nWe can do this using the second rule backwards:\n\nln (1/2) = ln 1 - ln 2\n\nand ln 1 is zero.\n\nOr we can do it using the third rule\n[remembering that 2^(-1) = 1/2].  \nHence:\n\nln (1/2) = ln (2^-1) = -1×ln 2\n\nHope that helps!
728396
1.\nln(5x + 2) + ln x = ln e\nln(5x + 2) + ln x = 1\ne ^ (ln(5x + 2) + ln x) = e ^ (1)\n(e ^ ln(5x + 2))(e ^ ln x) = e\n(5x + 2)(x) = e\n5x^2 + 2x - e = 0\nx = (-2 +/- sqrt(4 - (4)(5)(-e))) / 2(5)\nx = (-2 +/- sqrt(4 + 20e)) / 10\nx = (-2 +/- (2)sqrt(1 + 5e)) / 10\nx = (-1 +/- sqrt(1 + 5e)) / 5\ntest answers to ensure ln of positive number\nx = (-1 + sqrt(1 + 5e)) / 5\n\n2. \nln(x + 1) / ln(x - 1) = 2\nln(x + 1) = (2)ln(x - 1)\nln(x + 1) = ln((x - 1)^2)\nx + 1 = (x - 1) ^ 2\nx + 1 = x^2 - 2x + 1\n0 = x^2 - 3x\nx(x - 3) = 0\nx = 0 or x = 3\ntest answers to ensure ln of positive number\nx = 3
882612
2x^2 ln(x) - x^2\n\nVerify by differentiating\n2x + 4x ln(x) - 2x = 4x ln(x)
723369
ln e^(3x) = ln4\n3x(lne) = ln4  (note lne = 1)\n3x = ln4\nx = (ln4)/3\nx=0.462
754397
ln(x+3) - ln(x) = ln[(x+3)/3]\n\n2ln(2) = ln(2²) = ln4\n\nSo, if ln[(x+3)/x)] = ln4, then\n\n(x+3)/x = 4, or\n\nx+3 = 4x, or \n\nx = 3x, or x = 1\n\nI used the properties of logs to help me solve this.  I think you tried to make it too hard on yourself (which is very easy to do when dealing with the hated logarithms!).  Also, you brought e into it when you didn't really need to.  Since we had ln(x) = ln(y), essentially, then from there we can conclude that x=y without bringing e into play.  Anyhoo, here's the log properties I used in this solution.  (I use the word log but it applies to ln as well.):\n\nlog(a) - log(b) = log(a/b)\n\nrlog(a) = log(a^r)\n\nHope that helped!\n\n~ ♥ ~
935326
You have got to use expansion series!\n\nX->0+ [e-(1+x)^(1/x)]/x =e/2 \nlet x=0+h\n=>as x->0, h->o\n\nh->0 [e-(1+h)^(1/h)]/h\n\nlet y=(1+h)^(1/h)\n=>log y=(1/h)log(1+h)\n=>y=e^[(1/h)log(1+h)]\n\nexpansion of log(1+x)=x - (x^2)/2..................\n=>e^[(1/h)log(1+h)]= e^[(1/h)[h - (h^2)/2!............]\n\n=>e^[(1/h)[-h - (h^2)/2!............]=e^[1-h/2!........]\n\ntherefore,\n\nlet y =lim h->0 [e-(1+h)^(1/h)]/h=lim h->0 {[e- e^(1-h/2!...)]/h}\n\nexpansion of e^x=1+x+(x^2)/2!.................\n\n=>lim h->0 {[e- e^(1-h/2...)]/h}=lim h->0 {-e[e^(-h/2.....) -1]}/h\n\nexpansion of e^x=1+x+(x^2)/2!.................\n=>lim h->0{-e[1+(-h/2-(h^2)/3.....)+ ........-1]/h}\n\n=>lim h->0{-e[(-h/2-(h^2)/2...)+ [(-h/2-(h^2)/2...)^2]/2!+.....]}/h\n\n=>lim h->0{-e(-1/2 -..(terms containing highter powers of h))}\n\n=>e/2\n\n\nP.S.  By the way L' Hospitals rule would fail here as limit is not of form 0/0 or ~/~. It is of form a/0 where a is constant.
729730
u cant split a log like that! ie ln(ab) = lna +lnb\n\nu should use the f'(x) f(x)^n = f(x)^(n+1)/ (n+1)
793577
|e^x - 5| = e^x - 5 for e^x - 5 ≥ 0 ie x ≥ ln5\n= 5 - e^x for e^x - 5 < 0 ie x< ln5\n\nSo ∫|e^x - 5|dx from [-1,4]\n\n= ∫(5 - e^x) dx from [-1, ln5] + ∫(e^x - 5) dx ]\n\n= [5x - e^x] from [-1, ln5] + [e^x - 5x] from [ln 5,4\n\n= [(5ln5 - 5) - (-5 - e^(-1))] + [(e^4 - 20) - (5 - 5ln5)]\n\n= e^4 + e^(-1) - 15\n\n(≈ 34.97)
1012764
I seem to be the first with the correct solution... the confusion is obviously the notation.  If you can't type superscripts easily, you can always use ^ to mean exponentiation.  (Or Alt-253 on a PC keypad will get you a superscript ²).\n\nThe solution is to simply substitute x in your equation with whatever is put inside the parentheses.  It can be confusing if that includes another equation, but it is still simple substitution.\n\nGiven:\nf(x) = 2(x^2) -1 \n\nf(2x) = 2((2x)^2) - 1\nf(2x) = 2(4(x^2)) - 1\nf(2x) = 8(x^2) -1\nf(-x) = 2(-x^2) - 1\nf(-x) = 2(x^2) - 1\n\nf(2x) - f(-x) = 6(x^2)\n\nf(x+1) = 2((x+1)^2) - 1\nf(x+1) = 2(x^2 + 2x + 1) - 1\nf(x+1) = 2x^2 + 4x + 2 - 1\nf(x+1) = 2x^2 + 4x + 1\n\nSo your answers are:\n1) 6x^2\n2) 2x^2 + 4x + 1\n\nWritten using Alt-253 for the exponent, the answers are:\n1) 6x²\n2) 2x² + 4x + 1
234025
(m^(2n))/(3*(a^x)) = 37       \n(3a^x)(37) = m^(2n)            *****cross-multiply\n111a^x = m^(2n)                *****3*37=111\na^x = (m^(2n))/111             *****divide by 111\nx= log (base a) (m^(2n))/111   *****take log (base a) of both sides\n\nOne note: If you meant the entire expression (3a) to be raised to the x power, you need to use parentheses around it. Hope that helps.
933362
you should start with the power series representation of ln(1+x), because this is one of the standard ones that is either given or you have memorized\n\nln(1+x) = x - (1/2)x^2 + (1/3)x^3 - (1/4)x^4 + . . . \n\nso ln(4-x) is what you get when you sub in 3-x in for x in the above equation\n\nln(4-x) = (3-x) - (1/2)(3-x)^2 + (1/3)(3-x)^3 - (1/4)(3-x)^4 + . . . \n\nand now, xln(4-x) is when you multiply the entire series above by x\n\nxln(4-x) = x(3-x) - x(1/2)(3-x)^2 + x(1/3)(3-x)^3 - x(1/4)(3-x)^4 + . . .
829854
Use logarithms to bring exponents down where you can deal with them.  Since the natural log is base e, that's the way to go here.\n\ne^5x = 7\n\n5x = ln 7\n\nand solve from there.
226422
This one is trickier - ln-2/x wasn't allowed on the source. I tried the following:\n\nln[e]-2/x ==> 1-2/x ==> 2/x^2\nln[2/x] ==> -1/x\n\nI hope this helps.
739921
Not too hard.\n1. plug in what you know\n\nP/2 = Pe^(-0.0027x)\n\n2. cancel the Ps.\n1/2 = e^(-0.0027x)\n\n3. use some tricks with the ln and e to get x by itself\nln(1/2) = -0.0027x\n\n4. solve for x\nln(1/2) / (-0.0027) = x\n\n5. calculator, most definitely\nx = 256.72\n\n\nOn step 3, try to remember this:\na = e^b, therefore ln(a) = b. That's usually the make-or-break with exponential decays.