Express the given quantity as a single logarithm. ln(5) + 3 ln(3)?
Problem 1) .. ln(5) + 3 ln(3) = ln(5) + ln(3^3) = ln(5 * 27) = ln(135) Problem 2) .. ln(a + b) + ln(a - b) - 7 ln(c) = ln ( (a + b) * (a - b) ) - ln(c^7) = ln ( (a² - b² ) / c^7 )
Express the given quantity as a single logarithm: ln(1+x^5) + 1/2ln(x) - ln(sin(x))?
ln(1 + x^5) + ½ ln(x) - ln(sin(x)) = ln(1 + x^5) + ln( √x) - ln(sin(x)) = ln[ (1 + x^5)√x) / sin(x) ] you distributed the ½ incorrectly
Express the given quantity as a single logarithm.?
= ln(x) + ln(y^a) - ln(z^b) = ln[x(y^a)] / (z^b)
Express the given quantity as a single logarithm.?
Hi, ln(8) + 4 ln(3) = ln(8) + ln(3)⁴ = ln(8) + ln(81) = ln(8*81) = ln 648 <==ANSWER I hope that helps!! :-)
Express the given quantity as a single logarithm. 1/5ln(x +2)^5 + 1/2[lnx - ln(x^2 + 3x + 2)^2]?
(1 / 5) * ln[(x + 2)^5] + (1 / 2) * ln[(x) - ln(x^2 + 3x + 2)^2] First take the multiplier (1 / 5) and shoot it inside the logarithm as an exponent. Then apply the difference of logarithm properties to the quantities ln(x) - ln(x^2 + 3x + 2), where a = x, and b = x^2 + 3x + 2, in this property: ln(a) - ln(b) = ln(a / b). After you have formed your single logarithm, shoot the 1 / 2 multiplier as an exponent inside the logarithm: ln[(x + 2)^5^(1/5)] + {ln[(x / [(x^2 + 3x + 2)^2]^(1 / 2)} When a power is raised to a power multiply the powers, also factor the denominator in the second logarithm: ln(x + 2) + ln[√x / (x + 2)(x +1)}] Then apply the property: ln(a) + ln(b) = ln(a * b). = ln{(x + 2) * √x / (x + 2)(x+1)]} We can cancel an (x + 2) / (x + 2) expression to 1. = ln((√x / (x+1))
Express the given quantity as a single logarithm: 1/5 ln(x + 2)^5 + 1/2( ln(x) − ln(x^2 + 3x + 2)^2)?
if there is a plus in front of it, it will go in the numerator. If there is a minus sign infront of it it will go in the denominator. rules: alnx=lnx^a ln(x+2)+ 1/2 lnx -1/2ln(x^2+3x+2)^2 ln(x+2)+lnx^(1/2)-ln(x^2+3x+2) ln[(x+2)(x^(1/2)/(x^2+3x+2)]. I distributed the 1/2.
Please Help!........ Express the given quantity as a single logarithm ln(a + b) + ln (a - b) - 2 ln c?
Hi, ln(a + b) + ln (a - b) - 2 ln c = ln(a + b) + ln (a - b) - ln c² = . (a + b)(a - b) ln ------------------ = . . . . c² . . a² - b² ln ----------- <==ANSWER . . . c² I hope that helps!! :-)
Calculus question:Express the given quantity as a single logarithm.?
= ln((1+x^4)*x^(1/2)/cos(x))
How do you find the logarithm of a negative number? Why is it a complex number?
To answer the second part first, logarithms of positive numbers greater than one are positive, less than one have negative logarithms. So every real number is the logarithm of some positive number, so you can think of it as having "nothing left over" to be the logarithms of negative numbers. (This is to aid your understanding, it is NOT a rigorous argument.)OK. Let's take a simple case, -1.Now let's note that i is the square root of -1.We want to define logarithms of negative numbers to preserve the property we have for logarithms of positive ones. log(xy) = log(x) + log(y).Setting x=y=i:log(ii) = log(-1) = 2log(i).Since we see immediately that the log of a negative number will be a related to the log of an imaginary number in any base, we now strongly suspect that we need complex numbers here.Problem is, a logarithm is the opposite of an exponential function and we don't know how to do that with complex numbers. So we define it in terms of its power series, and substitute i, or whatever.It can now be shown from that that,[math]e^{i\pi} = -1[/math].This is called Euler's identity.Taking the natural log of both sides:[math]ln(e^{i\pi}) = ln(-1)[/math].[math]i\pi= ln(-1)[/math].So [math]i\pi = ln(-1)[/math].Once you have [math]ln(-1)[/math], you can calculate the natural log of any negative number.Let x be a positive real number. The negative -x, is the product of x and -1.Since we are preserving the property of a the log of a product being the sum of the logs.[math]ln(-x) = i\pi + ln(x)[/math].Nasty note: since the exponential function of complex numbers gives the same result for [math]z[/math] and [math]z+2n\pi[/math] for any integer n, the logarithm is actually a multivalued function. See Complex logarithm for details.