Wednesday

April 16, 2014

April 16, 2014

Number of results: 4,154

**Anti-ln?!?!**

use a calculator to find the anti-ln(-0.049) to the nearest ten thousandth. My calculator doesn't do anti-ln. just ln. Someone helo me please?
*Saturday, April 21, 2012 at 1:47pm by Tabby*

**math**

On my screen your intended symbol did not show up correctly. did you mean: ln e = ln (√3/(x) ) - 4ln e ? if so, then 5ln e = ln (√3/x) ln e^5 = ln (√3/x) anti-ln it e^5 = √3/x x = √3/e^5 Let me know if your equation was meant differently.
*Friday, March 15, 2013 at 10:04pm by Reiny*

**math**

ahh, my error, I thought it was question #4 easy to fix 4 ln e = ln (√3/(x) ) - 4ln e 8ln e = ln (√3/x) ln e^8 = ln (√3/x) anti-ln it e^8 = √3/x x = √3/e^8
*Friday, March 15, 2013 at 10:04pm by Reiny*

**Anti-ln?!?!**

Use the inverse key INV LN (-.049) or do this use the e^ key anti-ln(a)= e^a
*Saturday, April 21, 2012 at 1:47pm by bobpursley*

**math**

ln(a)-ln(b)=ln(a/b) ln(x+9)-ln(x)=1 ln[(x+9)/x]=1 anti-logarithm of that: (x+9)/x=e e=2.718281828459045235360287471 x+9=e*x x-e*x= -9 x*(1-e)= -9 x= -9/(1-e) x= -9/(1-2.718281828459045235360287471) x= -9/-1.718281828459045235360287471 x= 5.2377903618239378194650180459811
*Wednesday, March 16, 2011 at 6:33pm by Anonymous*

**Math**

Use the law of logarithms: ln(a)+ln(b)=ln(ab) ln(a)-ln(b)=ln(a/b) kln(a)= ln(ak) ln(1)=0 ln(1/x)=-ln(x) so ln(x)-ln(x²)+ln(5)=0 ln(x/x²)+ln(5)=0 ln(1/x)+ln(5)=0 -ln(x)+ln(5)=0 ln(x)=ln(5) e^(ln(x))=e^(ln(5)) x=5
*Wednesday, April 27, 2011 at 3:04pm by MathMate*

**precalc**

1. ln e^(ln x) - ln(x - 3) = ln 8 ----- apply the log rules (lnx)(lne) - ln(x-3) = ln 8 ----- recall that ln e = 1 lnx - ln(x-3) = ln 8 ln (x/(x-3)) = ln 8 ---- recall : ln (A/B) - ln A - ln B "anti-ln" it x/(x-3) = 8 cross-multiply 8x - 24 = x 7x = 24 x = 24/7 2. Divide both ...
*Friday, May 6, 2011 at 5:20pm by Reiny*

**math**

Actually I should have written it as ln{[x/(x^2-1]^2} The -ln(x+1)-ln(x-1) in the denominator is ln {1/[(x+1)(x-1)] = ln [1/(x^2-1)] Adding ln x puts the x in the numerator of [x/(x^2-1)] The two in front of 2[ln(x)-ln(x+1)-ln(x-1)] is the same as taking the log of the square ...
*Friday, January 23, 2009 at 10:18pm by drwls*

**algebra**

are we solving for x ?? if so, then ln[(x-5)(x+4)] = ln36 "anti-ln" it (x-5)(x+4) = 36 x^2 - x -56 = 0 (x-8)(x+7) = 0 x = 8 or x = -7 but x=-7 would make the original undefined, so x = 8
*Wednesday, May 12, 2010 at 1:59pm by Reiny*

**Math 2**

ln 4th root (x^6/(y^2 z)) I read that as ln (x^6/(y^2z)^(1/4) = (1/4) ( ln x^6 - ln y^2 - ln z) = (1/4)( 6lny - 2lny - lnz) = (3/2)lny - (1/2)lny - (1/4) lnz ln y^4 + (1/3) ln(x^9 y^9) − 7 ln y = ln y^4 + ln(x^9 y^9)^3 - ln y^7 = ln [y^4(x^3 y^3)/y^7] = ln (x^3) or 3lnx
*Wednesday, April 24, 2013 at 4:15pm by Reiny*

**Write the expression as a single log**

I am sure you meant to type ln( x/(x-5) ) + ln( (x+5)/x ) - ln(x^2 - 25) = lnx - ln(x-5) + ln(x+5) - lnx - (ln(x+5) + ln(x-5) = -ln(x-5) + ln(x+5) - ln(x+5) - ln(x-5) = -2ln(x-5)
*Sunday, April 7, 2013 at 9:20pm by Reiny*

**Math differentiation**

1. y = x^x 2. y = (sin^2xtan^4x)/(x^2+1)^2 3. y = x^(1/x) 4. y = 2^(-x^3) Okay, so I'm pretty sure 1 is done easily: ln y = ln (x^x) ln y = x ln x d/dx ln y = d/dx x ln x y'(1/y) = x(1/x) + ln x(1) = 1 + ln x y'(1/y) = ln x + 1 y' y(1/y) = (ln x + 1)y y' = (ln x + 1)y since y...
*Wednesday, August 27, 2008 at 1:52am by Marissa*

**College Algebra**

1. fog(x) = 2 /[(7/x) - 3] simplifying, 2x/(7 - 3x) Domain: all real numbers except 3, 0 and 7/3 2. 2^(1-9x) = e^(2x) get ln of both sides: ln 2^(1-9x) = ln e^(2x) (1-9x)*(ln 2) = 2x (1-9x)/(2x) = 1/(ln 2) 1/(2x) - 9/2 = 1/(ln 2) 1/(2x) = 1/(ln 2) + 9/2 2x = 1/[1/(ln 2) + 4.5...
*Tuesday, August 14, 2012 at 1:45pm by Jai*

**pre calculus**

ln(x-5/x^2 - 1) - ln (x^2 - 2x -15/x + 1) = ln(x-5) - ln(x^2 - 1) - (ln(x-5)(x+3) - ln(x+1) ) = ln(x-5) -( (ln (x+1) + ln(x-1) ) - ( ln(x-5) + ln(x+3) - ln(x+1) ) = ln(x-5) - ln(x+1) - ln(x-1) - ln(x-5) - ln(x+3) + ln(x+1) = -( ln(x-1) + ln(x+3) ) = -ln( (x-1)(x+3) ) or -ln(x^...
*Saturday, December 17, 2011 at 9:15pm by Reiny*

**CALCULUS**

Let u = ln x d/dx ln ln x = d/dx (ln u) = dln u/du * du/dx = (1/u)* (1/x) = 1/(x ln x) Now add another step. Let v = ln (ln x) d/dx ln ln ln x = d/dx ln v = d(ln v)/dv * dv/dx = (1/v)/(x ln x) = [1/(ln ln x)]*1/(x ln x) = 1/[x ln x]^2 In order for the function to be defined, x...
*Thursday, October 25, 2007 at 8:32pm by drwls*

**Calculus (Thanks a bunch!)**

Since ln(a)+ln(b) = ln(ab) ln(a)-ln(b) = ln(a/b) (1/2)ln(a) = ln(sqrt(a), and 3ln(a) = ln(a3) 5 ln(x) = ln(x5) -(1/2)ln(y) = ln(1/sqrt(y)) 3 ln(z) = ln(z3) So 5lnx - 1/2 lny + 3 lnz =ln(x5 * 1/sqrt(y) * ... ) Can you now complete the answer and simplify?
*Friday, June 26, 2009 at 4:05pm by MathMate*

**Calculus**

ln ( x ^ 4 ) = 4 * ln ( x ) ln ( x ^ 2 ) = 2 * ln ( x ) ln ( x ^ 4 ) - ln ( x ^ 2 ) = 2 4 * ln ( x ) - 2 * ln ( x ) = 2 2 * ln ( x ) = 2 Divide both sides by 2 ln ( x ) = 1 Cancel logarithm by taking exp of both sidess x = e = 2.71828...
*Monday, September 24, 2012 at 12:30am by Bosnian*

**exponential equation-Algebra**

Take the natural logs (ln) of both sides. (x+4) ln 4 = (2x+5) ln 5 x ln 4 + 4 ln 4 = 2x ln 5 + 5 ln 5 x (2 ln 5 - ln 4) = 4 ln 4 - 5 ln 5 x = (4ln4 - 5ln5)/(2ln5 - ln4) This becomes the same as your answer if you multiply both numerator and denominiator by -1
*Friday, October 26, 2007 at 10:59pm by drwls*

**Algebra**

put to work the properties of logs. ln x + ln(x+1) = 1 ln(x(x+1)) = 1 x(x+1) = e x^2 + x - e = 0 then use the quadratic formula ln(x+5) = ln(x-1)/ln(x+1) ln(x+5) = ln((x-1)/(x+1)) x+5 = (x-1)/(x+1) x^2+6x+5 = x-1 x^2+5x+6 = 0 ...
*Wednesday, October 30, 2013 at 10:40pm by Steve*

**Math: Calculus**

by definition, 2 = e^(ln 2) Therefore, 2^t = e^(ln 2)^t = e^(t ln 2) Make sense yet? Now, ln (2^t) = t * ln 2 So, ln N = ln 40 + t ln 2
*Monday, October 17, 2011 at 7:37pm by Steve*

**Anti-ln?!?!**

Tabby, I had a Casio at one time Look for a 2ndF key, it works the same way enter: 2ndF ln -.049 =
*Saturday, April 21, 2012 at 1:47pm by Reiny*

**Calculus**

2^3 take ln and get ln(2^3) = 3 ln 2 3^ln 2 take ln and get ln (3^ln 2) = ln 2 ln 3 2^ln 3 also gives ln 3 ln2 and ln 3 <3
*Sunday, October 3, 2010 at 7:45pm by Damon*

**MATHHH**

log2(8)=3 log9(3)=0.5 So, (3)^x = (0.5)^(x+1) ln of both sides (x)(ln(3) = (x+1)(ln(0.5) (ln(3) - ln(0.5))(x)= ln(0.5) calculate ln(3) and ln(0.5) solve for x
*Sunday, December 13, 2009 at 10:06pm by Quidditch*

**Math**

ln x^2(x+1)/x+2 Write in terms of ln(x), ln(x+1), and ln(x+2) is the correct steps this? lnx^2 + ln(x+1) - ln(x+2) = 2lnx + ln(x+1) - ln(x+2)
*Thursday, September 29, 2011 at 7:38pm by Blue*

**calculus**

Note that f'(x) can be rewritten algebraically a f'(x)= x -(1/x). Now, take the integral (anti-drivitive) of f'(x) with respect to 'x' to obtain f(x). So, Int(x-(1/x),x)=((x^(2))/2)-ln(abs(x))+C To understand this you must first look at the rules of integrals. By the way in ...
*Sunday, November 8, 2009 at 6:10pm by Collin*

**Help, Math.**

Solve: 3e2x + 2 = 50 a. ln 8 b. ln 0.25 c. ln 14 d. 0.5ln 16 --------- Solve: 1 = 7 ex a. ln 1/7 b. ln 1/7 c. ln 7 d. ln 7 ------------ Solve: e3x = 8 a. ln 5 b. 3 ln 8 c. ln 8 3 d. ln 3 8 ---------- ln 1 √ e a. -√e b. -1 2 c. 1 e d. 1 2
*Tuesday, May 21, 2013 at 3:49pm by Holly*

**math**

You MUST put brackets for the denominators to avoid ambiguity I am sure you mean ln ( x/(x+1) ) + ln( (x+1)/x ) - ln (x^2 - 1) then = lnx - ln(x+1) + (ln(x+1) - lnx - ln( (x+1)(x-1) = lnx - ln(x+1) + ln(x+1) + lnx - ln(x+1) - ln(x-1) = - 2ln(x+1) or ln [ (x/(x-1) ((x+1)/x...
*Thursday, January 3, 2013 at 11:46pm by Reiny*

**Math**

3 ln 2 – ln 4 + 5 ln x = ln 3 + 2 ln 5 ln 2^3 - ln4 + ln x^5 = ln 3 + ln 5^2 ln ((8x^5)/4) = ln (75) none of those match this left in log form, I had ln (8x^5/4) = ln 75 none of your choices match this.
*Sunday, February 16, 2014 at 6:59pm by Reiny*

**Math**

In the theorem: ln e^x = x * ln e = x put x = ln(y): ln{exp[ln(y)]} = ln(y) Then we have: exp(ln(y)) = y if it is allowed to conclude from ln(a) = ln(b) that a = b This follows from the definition of ln: ln x = definite integral from 1 to x of 1/t dt Since (for positive x), 1/...
*Thursday, March 12, 2009 at 12:39pm by Count Iblis*

**MATH HELP!**

n * ln ( a ) = ln ( a ^ n ) 2 ln ( x ) = ln ( x ^ 2 ) 3 ln ( y ) = ln ( y ^ 3 ) ( 1 / 2 ) ln ( z ) = ln [ z ^ ( 1 / 2 ) ] = ln [ sqrt ( z) ] ln ( a ) + ln ( b ) = ln ( a * b ) ln ( a ) - ln ( b ) = ln ( a / b ) In this case: 2 ln ( x ) - 3 ln ( y ) + ( 1 / 2 ) ln ( z ) = 2 ln...
*Sunday, April 1, 2012 at 10:20pm by Bosnian*

**Calculus**

is that ln (x^2) or (ln x)^2 ? anyway, in any case, it is not differentiable at x = 0 because if you get the derivative, for ln (x^2): ln (x^2) = 2*ln x derivative of 2*ln x = 2/x for (ln x)^2: derivative (ln x)^2 = 2*(ln x)/x if you substitute x=0 to the derivative, the ...
*Friday, October 29, 2010 at 12:42am by jai*

**Math!!!**

I am not sure if your 10^3 is in the numerator or denominator so will assume it is in the denominator/ ln(a/b) = ln a - ln b so ln x - ln (5 * 10^3) = .329 but ln (a * b) = ln a + ln b so ln x - ln 5 - ln 10^3 = .329 but ln (a^b) = b lan a so ln x - ln 5 - 3 ln 10 = .329 or ln...
*Monday, June 16, 2008 at 2:50pm by Damon*

**Math**

Which equation is equivalent to 3 ln 2 – ln 4 + 5 ln x = ln 3 + 2 ln 5? ln 2x5 = ln 75 ln (5x + 2) = ln 75 ln 2x5 = ln 13 ln (5x + 4) = ln 96
*Sunday, February 16, 2014 at 6:59pm by Kiki*

**Math**

Which equation is equivalent to 3 ln 2 – ln 4 + 5 ln x = ln 3 + 2 ln 5? ln 2x5 = ln 75 ln (5x + 2) = ln 75 ln 2x5 = ln 13 ln (5x + 4) = ln 96
*Sunday, February 16, 2014 at 8:32pm by Kiki*

**AP Algebra**

ln ( a ^ n ) = n * ln ( a ) ln ( a ) + ln ( b ) = ln ( a * b ) ln ( a ) - ln ( b ) = ln ( a / b ) 2 ln ( x ) + 3 ln ( y ) - ln ( 2 ) - 4 ln ( z ) = 2 ln ( x ) + 3 ln ( y ) - [ ln ( 2 ) + 4 ln ( z ) ] = ln ( x ^ 2 * y ^ 3) - ln ( 2 z ^ 4 ) = ln ( x ^ 2 * y ^ 3 / 2 z ^ 4 )
*Tuesday, December 3, 2013 at 3:31pm by Bosnian*

**Algebra**

ln e^lnx-ln(x-3)=ln8 First of all the "ln of e" cancels out, so it is just: ln(x)-ln(x-3)=ln(8) which is the same as: ln(x/(x-3))=ln(8) e^ both sides... x/(x-3)=8 x=8x-24 7x=24 x=24/7
*Saturday, April 12, 2008 at 6:32pm by Bobby*

**amy**

Use the laws of logarithm: ln a + ln b = ln ab 2ln a = ln a² 2 ln(x-3)= ln(x+5) + ln4 => ln(x-3)² = ln 4(x+5) Raise to power of e: (x-3)² = 4(x+5) Simplify and factor to get: (x-11)*(x+1)=0 Can you take it from here?
*Monday, March 14, 2011 at 10:00pm by MathMate*

**math**

Express y as a function of x. C is a positive number. 3 ln y= 1/2ln (2x+1) - 1/3ln (x+4) + ln C ln y = (1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C y = e^ln y = e^[(1/6)ln (2x+1) -(1/9)ln (x+4) + (1/3) ln C] = C^(1/3)* (2x+1)^(1/6) /(x+4)^(1/9)
*Wednesday, January 10, 2007 at 12:18am by Denise*

**calculus**

remember that log(product) = sum of logs of factors let y = 2^x(x^2+2)^3(x^3-3)^7/(x^2+4)^1/2 take ln of both sides ln y = ln 2^x + ln (x^2+2)^3 + ln (x^3 - 3)^7 - ln (x^2+4)^(1/2) ln y = x (ln2) + 3 ln(x^2+2) + 7 ln(x^3-3) - (1/2) ln(x^2 + 4) y' /y = ln2 + 6x/(x^2 + 2) + 21x^...
*Monday, April 29, 2013 at 8:40pm by Reiny*

**Math - Write expression as single logarithm**

ln((x^2 – 1)/(x^2 – 6x + 8)) – ln((x+1)/(x+2)) when you subtract the logs, you divide the arguments ln (x+1)(x-1) / [(x+1)(x-1)((x-4)(x-2)] ln 1/[(x-4)(x-2)] ln [ (x-4)(x-2) ]^-1 -ln [ (x-4)(x-2)]
*Monday, June 6, 2011 at 1:44pm by Damon*

**Math**

Given the axioms: ln x = definite integral from 1 to x of 1/t dt ln e = 1 Given the theorems: ln a^r = r * ln a (for all real numbers r) ln e^x = x * ln e = x Prove that e^(ln x) = x Intuitively, I can see that is true, but how can it be proved?
*Thursday, March 12, 2009 at 12:39pm by Sean*

**Logarithms**

Use the product rule, ln(a)+ln(b) = ln(ab) ln(a)-ln(b) = ln(a/b) Work out the problem and post your answer for a check if you wish.
*Friday, June 25, 2010 at 12:57pm by MathMate*

**Anti-ln?!?!**

Thanks Reiny!
*Saturday, April 21, 2012 at 1:47pm by Some1 Help Me Plz*

**biology**

Do production of anti-A, anti-B, and anti-Rh antibodies require the exposure to antigens
*Tuesday, September 11, 2012 at 4:54pm by amy*

**Calculus I**

first of all recall that ln (A/B) = ln A - ln B so y = ln((1+e^x)/(1-e^x) ) = ln (1+e^x) - ln (1-e^x) dy/dx = e^x/(1+e^x) - (-e^x)/(1 - e^x) simplify as needed
*Monday, May 21, 2012 at 12:18pm by Reiny*

**Math**

What does the following infinite series starting at k=2 converge to: Σ ln (1 - 1/k^2) In other words, what does this converge to: ln(1 - 1/4) + ln(1 - 1/9) + ln(1 - 1/16) + ln(1 - 1/25) + ln(1 - 1/36) + ... I assume the first step is this: Σ ln (1 - 1/k^2) = ln &Pi...
*Thursday, May 28, 2009 at 12:28pm by Sean*

**pre-calculus**

Assuming you meant: In(x/(x-5))+In((x+5)/x)-In(x^2-25) = lnx - ln(x-5) + ln(x+5) - lnx - ( ln(x+5) + ln(x-5) ) = lnx - ln(x-5) + ln(x+5) - lnx - ln(x+5) - ln(x-5) = -2ln(x-5) or -ln(x-5)^2 or ln(x-5)^-2
*Tuesday, October 9, 2012 at 9:39pm by Reiny*

**Math (pre calculus)**

if so then (3x-1) ln 2 = (2x+1) ln 5 3 x ln 2 - 2 x ln 5 = ln 5 + ln 2 x (ln 8 - ln 25) = ln 10 x ln (8/25) = ln 10 x = ln 10 / ln (8/25)
*Thursday, December 12, 2013 at 6:48pm by Damon*

**math**

(e^(e^x)) = 2 ln both sides ln( (e^(e^x))) = ln 2 e^x(ln e) = ln 2, remember ln e = 1 e^x = ln2 ln both sides again, and by the same concept x = ln(ln 2))
*Saturday, August 14, 2010 at 6:01pm by Reiny*

**Math**

3 ln 2 – ln 4 + 5 ln x = ln 3 + 2 ln 5 ln (2^3 / 4 * x^5) = ln(3 * 5^2) ln 2x^5 = ln 75
*Sunday, February 16, 2014 at 8:32pm by Steve*

**Math**

For r=0.095, with normal compound interest, compounded yearly, the number of years to double at a rate of r% is ln(2)/ln(1+r)=ln(2)/ln(1.095)=7.638 years. from A=P(1+r)^n take ln both sides, ln(A/P)=n ln(1+r) n=ln(2)/ln(1+r) With continuous compounding, A=Pern Take logs ln(A/P...
*Sunday, November 21, 2010 at 8:09pm by MathMate*

**algebra**

27 = x ^ -3 3 ^ 3 = 1 / (x^3) {factorise 27 and invert x^-3} ln(3 ^3) = ln( 1 / (x^3) ) { ln both sides} ln(3^3) = ln(1) - ln(x^3) {apply the log rule for division} 3ln(3) = 0 - 3ln(x) {ln(1) = 0 in all bases} ln(3) = -ln(x) -ln(3) = ln(x) {multiple both sides by -1} ln(3 ^ -1...
*Friday, March 1, 2013 at 5:17pm by Anonymous*

**Sequence and series**

g = a, ar, ar^2, ar^3, ... ln g = ln(a), ln(a) + ln(r), ln(a) + 2ln(r), ln(a) + 3ln(r), ... so, the AS has first term = ln(a) and difference = ln(r)
*Saturday, January 21, 2012 at 6:26pm by Steve*

**math**

8 ln x -(1/3) ln y = ln(x^8) - ln(y^(1/3)) = ln (x^8/(y^(1/3)) if in b) you mean ln(x^8/cuberoot(y)) that would be it, and the same as mine
*Sunday, December 6, 2009 at 11:18am by Reiny*

**Math - Calculus**

except that page teaches you nothing, what are you going to do on a test? g(x) = ln[(x^3+1)^3(x^3-1)^3)] = 3 ln(x^3 + 1) + 3 ln (x^3 - 1) g'(x) = 3(3x^2)/(x^3+1) + 3(3x^2)/(x^3 - 1) = 9x^2/(x^3+1) + 9x^2/(x^3-1) The Wolfram page assumed that you meant ln another way is to ...
*Monday, November 14, 2011 at 1:10pm by Reiny*

**Algebra 2**

Using quotient rule of logarithms ln(m/n) = ln(m) – ln(n) so ln(1)-ln(2) = ln(1/2)
*Friday, May 7, 2010 at 8:57pm by Chris*

**calculus**

I should have had 4x^4 = 4^x ln 4 x^4 = ln 4 + 4 ln x so ln 4 + 4 ln x = x ln 4
*Saturday, October 8, 2011 at 6:41pm by Damon*

**precalculus**

first x^2+12x+36 = (x+6)(x+6) so ln of (x-7)(x+6) ===================== (x+6)(x+6)(x+7)(x-7) ln of 1 ===================== (x+6)(x+1) = ln 1 - ln(x^2 +7x + 6) but ln 1 = 0 so - ln (x^2 + 7 x + 6)
*Wednesday, January 11, 2012 at 7:28pm by Damon*

**calculus**

remember that ln(a) + ln(b)=ln (ab) and a ln b = ln b^a and ln c - ln d= ln (c/d)
*Tuesday, June 30, 2009 at 4:43pm by bobpursley*

**Algebra**

ln(x^2+1)^(2/3)+ln((X^2-1)^(1/2)/(x+1)^(1/2)) = ln(x^2+1)^(2/3)+ ln((x+1)(x-1))^(1/2)/ (x+1)^(1/2)) = ln(x^2+1)^(2/3)+ln(x-1)^(1/2) = ln((x^2+1)^(2/3)*(x-1)^(1/2)). OR ln((crt(x^2+1)^2*sqrt(x-1).
*Monday, August 22, 2011 at 2:56pm by Henry*

**Pre-Calculus**

I need help solving this logarithmic equation: ln x + ln (x+3) = 1. I know ln x + ln (x+3)=ln [x(x+3)] and that is ln x^2 + 3x when broken down. Then you have to get rid of ln by taking the natural base e of both sides so in the end you have x^2+3x = e^(1). That's when I got ...
*Saturday, May 3, 2008 at 11:05pm by Kelsey*

**Cal**

ln(0.9) = -0.1054 ln(1.0) = 0 ln(1.1) = 0.0953 (.0953 + 0.1054)/0.2 = 1.0035 ln(1.9) = 0.6419 ln(2.0) = 0.6931 ln(2.1) = 0.7419 (0.7419 - 0.6419)/0.2 = 0.5000 ln(4.9) = 1.5892 ln(5.0) = 1.6094 ln(5.1) = 1.6292 (1.6292 - 1.5892)/0.2 = 0.2000 ln(9.9) = 2.2925 ln(10.0) = 2.3026 ...
*Monday, September 26, 2011 at 12:30am by Steve*

**Anti-ln?!?!**

correct. Put this in your google search window: e^-.049=
*Saturday, April 21, 2012 at 1:47pm by bobpursley*

**Anti-ln?!?!**

Ah. Thank syou o much for your time and help. God Bless you.
*Saturday, April 21, 2012 at 1:47pm by Tabby*

**ALGEBRA**

Contract the expressions. That is, use the properties of logarithms to write each expression as a single logarithm with a coefficient of 1. text ((a) ) ln\(3\)-2ln\(4\)+ln\(8\) ((b) ln\(3\)-2ln\(4+8\) (c) )ln\(3\)-2(ln\(4\)+ln\(8\))
*Monday, April 16, 2012 at 10:48am by Randy*

**algebra 2**

original function: y = e^-x + 5 inverse function: x = e^-y + 5 e^-y = x-5 ln (e^-y) = ln(x-5) -y = ln(x-5) y = -ln(x-5) f^-1(x) = -ln(x-5) do the same for your second question, but make sure you use log instead of ln
*Thursday, July 14, 2011 at 9:23am by Reiny*

**Calculus**

That does not look like a function to be integrated. It looks like a number to be calculated. You can start out simplifying it by recognizing that ln(1) = 0. When you write ln(3)^2, do you mean [ln(3)]^2 or ln[3^2]? They are not the same. I will assume the latter. ln [3^2] = 2...
*Sunday, March 30, 2008 at 9:51pm by drwls*

**Calculus**

I'm confused. When I wrote it out by hand, I got: ð(3ln(9)-6ln(3)+6)-(ln(1)-2ln(1)+2) ð(ln(93)-ln(36)+6)- 2 (since ln(1)=0) ð(ln(729)-ln(729)+6)-2 6ð-2
*Sunday, March 30, 2008 at 9:51pm by Lin*

**Calculus derivative**

recall that if y = e^u where u is a function of x, y' = e^u du/dx Now, a^x = (e^(ln a))^x = e^(x*ln a) so, y' = e^(x*ln a) * ln a = ln a * a^x makes sense, since if a>e, ln a > 1, and the curve rises more steeply
*Thursday, September 27, 2012 at 8:31am by Steve*

**Maths**

Expand ln(1+x) and ln(1-x) and hence deduce that ln((y+1)/(y-1))=[2/y]+[2/(3(y^3))]+[2/(5(y^5))]+[2/(7(y^7))]. State the range of values of y for which the expansion is valid. Hence calculate ln(101) correct to 4 significant figures given that ln(99)=4.59507. I managed to ...
*Wednesday, October 22, 2008 at 9:38am by Claire*

**Calculus**

2. Express In 0.25 in terms of In 2 and In 3. Simplify your answer. Use integers or fractions for any number in the answer. ln (1/4) = ln 1 - ln 4 = 0 - ln 4 -ln 4 = -2 ln 2
*Saturday, January 16, 2010 at 7:45pm by Damon*

**Calculus**

I guess maybe you mean y = ln 3x^2 ???? y = ln 3 x^2 = ln 3 + ln x^2 = ln 3 + 2 ln x so dy/dx = 0 + 2 d/dx (ln x) = 2/x
*Friday, July 25, 2008 at 11:45am by Damon*

**pre cal**

(1/3)LN27 + 2LNx = LN(2-x) i think i just need help getting started. I assume that your LN notation is log to base e. I will call it ln. (1/3)ln 27 = ln (27)^(1/3) = ln 3, (since 27^(1/3) is the cube root of 27). Therefore ln 3 + 2 ln x = ln (3 x^2) = ln (2-x) 3x^2 = 2-x solve...
*Sunday, February 25, 2007 at 5:48pm by Marie*

**help plz math**

ln ( a ) - ln ( b ) = ln ( a / b ) ln ( 3 x + 5 ) - ln ( 4 ) = ln [ ( 3 x + 5 ) / 4 ] ln ( 3 x + 5 )- ln ( 4 ) = ln [ ( 3 x + 5 ) / 4 ] = ln ( x + 1 ) Cancel logarithms by taking exp of both sides ( 3 x + 5 ) / 4 = x - 1 Multiply both sides by 4 3 x + 5 = 4 * ( x - 1 ) 3 x + 5...
*Thursday, May 3, 2012 at 6:50pm by Bosnian*

**Calc**

Use chain rule. Let y=6xln(x) take log on both sides: ln(y) = ln(6)+ln(x)*ln(x) ln(y) = ln(6) + ln²(x) differentiate both sides w.r.t. x (1/y)y' = 0 + 2ln(x)*(1/x) y' = dy/dx = y*2ln(x)*(1/x) = 6xln(x)*2ln(x)*(1/x) = 12xln(x)-1*ln(x) Check my calculations.
*Thursday, November 4, 2010 at 12:24am by MathMate*

**college algebra**

Basic rules: lna+lnb=ln(ab) and lna-lnb=ln(a/b) also lna^x=xlna ln[x/(x-2)*(x+2/x)]-ln(x^2-4)= ln[(x^2+2)/(x-2)]-ln(x^2-4)= ln [divide]hence simplify fraction and get answer
*Tuesday, November 27, 2012 at 11:57pm by kay*

**CALCULUS PLEASE HELP!!**

1. .5 = e^1.5k ln .5 = 1.5 k ln.5÷1.5 = k m(4) = 400e^4(ln.5÷1.5)=62.996 mg m(8) = 400e^8(ln.5÷1.5)=9.921 mg 2. If half life =t1/2 .5 = e^kt1/2 ln.5 = kt1/2 k=ln.5/t1/2 hr 3. .01 = e^t(ln.5÷1.5) ln.01 =t( ln.5 ÷1.5) ln .01/( ln.5 ÷1.5)=t 9.966 hr=t
*Sunday, April 15, 2012 at 1:53pm by ####Nadine####*

**AP Algebra**

2lnx+3lny-ln2-4lnz ln(x^2) + ln(y^3) - ln(2) - ln(z^4) ln(x^2y^3 / 2z^4)
*Tuesday, December 3, 2013 at 3:31pm by Steve*

**algebra**

can someone plzs help me solve the equations by finding the exact solutions ln(x)-ln(5)=7 the next one is ln(e)=ln(with a v like symbol with 2/x)-ln(e)
*Sunday, October 28, 2012 at 3:48pm by missfubu-c*

**math**

without some grouping symbols, I know what you wrote. my guess is you meant this 2*ln(e)= ln ((7/x)^.5 ) - 2ln(e) 2*1=1/2 ln (7/x) -2 4=1/2 ln(7/x) ln(7/x)=8 take the antilog 7/x=e^8 then solve for x
*Sunday, November 10, 2013 at 1:08pm by bobpursley*

**alg 2.**

ln 25 = ln 5^2 = 2 ln 5 so we have 2 ln 5 - 1 ln 5 which is one ln 5
*Thursday, May 8, 2008 at 8:30pm by Damon*

**Calculus AB**

What you do with ln(xy) is just separate it into 2 pieces: ln(x) + ln(y) ln(x) + ln(y) = 1 + y' 1/x + y'/y = 1 + y' 1/x - 1 = y' (1 - y)
*Tuesday, February 21, 2012 at 12:02pm by Raf*

**calculus**

The first derivative of ln (ln x) is dy/dx = 1/(x*ln x) The second derivative is: d^2y/dx^2 = [-ln x/x^2 - (1/x^2)]/(ln x)^2 = -1/(x^2 ln x) -1/(x ln x)^2 x d^2y/dx^2 = -1/(x ln x) - (1/x)/lnx^2 It seems to fit the differential equation
*Sunday, March 30, 2008 at 1:40am by drwls*

**Calculus**

are we taking ln of the whole thing?? I read it as d( ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ] /dx let y = ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ] remember ln(AB) = lnA + lnB so our y = ln (x^3)^(1/2) + ln (x+3)^(1/3) + ln (3x-2)^(1/5) and remember ln A^n = n lnA = 3(1/...
*Saturday, December 1, 2012 at 9:49am by Reiny*

**math**

y = 3x - 2(4^x) Yep, the derivative of 3x is 3. Yep, it's true that derivative of 2 is 0, but this is not necessary. Note that 2 is multiplied by 4^x, and does not act as a separate term. Yep, the derivative of 4^x is 4^x(ln(4)). Therefore, y' = 3 - 2*(4^x (ln(4))) or note ...
*Tuesday, October 15, 2013 at 2:40pm by Jai*

**Math**

Your start was incorrect... the exponent does not apply to the 10, only to the base of e so ... e^(3x-7) = 5/10= .5 now take ln of both sides (3x-7) ln e = ln .5 , but ln e = 1 3x-7 = ln .5 3x = 7 + ln .5 x = (7 + ln .5)/3 , and it is button-pushing time
*Monday, April 9, 2012 at 1:57am by Reiny*

**Calculus/Logarithms**

y = (x^.5) ln x dy/dx = (x^.5) /x + ln x (.5 x^-.5) but x^.5 / x = x^-.5 so = (x^-.5) (1+.5 ln x) = (1+.5 ln x) / x^.5 = (2+1 ln x)/2 x^.5 so I agree with the book.
*Monday, June 29, 2009 at 8:21pm by Damon*

**math**

Nope, not right. try these rules... a ln b= ln( b^a) and ln c + ln d= ln (cd).
*Tuesday, July 28, 2009 at 1:46pm by bobpursley*

**algebra**

ln e = 1 , so your equation becomes 1 = ln(√(2/x) - 1 2 = ln √(2/x) √(2/x) = e^2 square both sides 2/x = e^4 x = 2/e^4 notice how I read your expression If you meant ln e = ln (√2 /x) - ln 2 2 = ln (√2/x) √2/x = e^2 x = √2/e^2
*Saturday, November 3, 2012 at 9:54pm by Reiny*

**Algebra**

e^x and e^2x x = 0, e^0 = 1 e^2*0 = 1 x = 1, e^1 = 2.72, e^2 = 7.39 x = 2, e^2 = 2.72, e^4 = 54.6 x = 3, e^3 = 20.1, e^6 = 403 etc (diverge fast from each other) ln x and ln 2x do not try ln 0 x = 1 , ln 1 = 0, ln 2 = .693 x = 2 , ln 2 = .692 , ln 4 = 1.39 x = 3 , ln 3 = 1.1...
*Monday, February 16, 2009 at 3:04pm by Damon*

**more of the same**

yes, you do get off track. Line 3 should be (because ln ab = ln a + ln b) ln 230=ln 305 + ln( e^-10*k) ln 230 - ln 305 = -10k because ln(e^x) = x now go from there. In fact, I'd have divided first: 230=305 e^-10*k 230/305 = e^-10k
*Monday, March 4, 2013 at 5:27pm by Steve*

**calculus**

Hi there, lim [ln(x-1)-ln(x+1)] b, -2 cannot get (ln(3)-0)-(oo-oo)) because ln(-00) has no definition. We should make it ln(x-1)-ln(x+1)=ln[(x-1)/(x+1)],then lim[ln[(x-1)/(x+1)]]b,-2 b->-00 is ln(3).
*Monday, February 11, 2008 at 3:41pm by H*

**pre-cal**

I assume you meant ln 2 ln(x+3) - 3 lnx = ln(x+3)^2 - ln x^3 = ln [ (x+3)^2/x^3 ]
*Tuesday, October 9, 2012 at 9:47pm by Reiny*

**Algebra**

how would you solve these two equations. ln x + ln(x+1)=1 ln(x+5) = ln(x-1) - ln(x+1)
*Wednesday, October 30, 2013 at 10:40pm by Kyle*

**Calculus**

I would split it into two problems and add the result. y = 5^x ln y = x ln 5 y'/y = ln 5 y' = y ln 5 y' = 5^x ln 5 then y = 15 x^-6 y' = 15(-6) x^-7 y' = -90/x^7 so finally y' = 5^x ln 5 - 90/x^7
*Monday, July 18, 2011 at 4:37pm by Damon*

**Math**

"ln" is term that designates the natural (base e) logarithm of whatever follows. ln x is the integral of dx/x ln (q+v) is the integral of dv/(q+v), with q being a constant. When you do the integration, you get the ln(q+v) - ln (q-v). Using the rules of logarithms, this can be ...
*Saturday, December 8, 2007 at 11:49pm by drwls*

**Calculus - derivatives**

let u=xx ln(u) = x ln(x) (1/u)du/dx = ln(x) + 1 du/dx = u (ln(x) + 1) du/dx = xx(ln(x) + 1) So this confirms what you've got for d(xx)/dx To find (x^x)^(x^x), it is not as complicated as it looks once we've got the derivative of xx. I would start with u=xx and du/dx = xx(ln(x...
*Sunday, November 29, 2009 at 3:58am by MathMate*

**math**

Integration of x^2 ln(x^3+2) | = integration symbol | x^2 ln(x^3 + 2) w = x^3 + 2 dw = 3x^2 1/3 dw = x^2 1/3 | ln(w) dw Integration by parts | u dv = uv - | v du u = ln(w) dv = dw du = 1/w dw v = w 1/3 | ln(w) dw = 1/3 ln(w) w - 1/3 | w 1/w dw = 1/3 w ln(w) - 1/3 | w/w dw = 1/...
*Wednesday, February 23, 2011 at 4:05pm by Helper*

**Quick Math Question**

3 ln x = ln x^3 ln x^3 + ln|x+1| = ln [x^3(x+1)] = ln [x^4 + x^3]
*Wednesday, March 26, 2014 at 6:23am by Damon*

**math**

you will have to use natural logs e^-7x = 2.5 ln (e^-7x) = ln 2.5 -7x(ln w) = ln 2.5 -7x = ln 2.5 , since ln e = 1 x = ln2.5/-7 = appr. -.1309
*Sunday, November 7, 2010 at 8:52am by Reiny*

Pages: **1** | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>