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The ln(y) is only defined for positive values of y.
So, ln(x2 + 2x - 8) is defined only for x2 + 2x - 8 > 0
ie for x< -4 or x > 2
So the domain is x< -4 or [union] x > 2
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How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Whai is the solution to Ln 2x plus 9 equals ln 5x?
That all depends.Is it:ln (2x+9) = ln (5x)If it is, the solution can be found by solving 2x+9 = 5x, in which case the answer is x = 3.
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
What is the solution set of 2.5 to the power of 2x equals 8?
2.52x = 8 Taking (natural) logs: 2x*ln(2.5) = ln(8) so 2x = ln(8)/ln(2.5) = 2.0794/0.9163 = 2.2694 approx. and therefore, x = 2.2694/2 = 1.1347 approx.
What is the value of x given the equation 5 to the power of 1- 2 x equals 0.25 correct to 3 decimal places Please show working out?
51-2x = 0.25 51 * 5-2x = 0.25 5-2x = 0.05 -2x*ln(5) = ln(0.05) x = ln(0.05)/[-2*ln(5)] = 0.931
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Whai is the solution to Ln 2x plus 9 equals ln 5x?
That all depends.Is it:ln (2x+9) = ln (5x)If it is, the solution can be found by solving 2x+9 = 5x, in which case the answer is x = 3.
What is the integral of -2x divided by the square root of x squared minus 4?
- ln ((x^2)-4)
Is 5 raised to 2x is equal to 18?
5(2X) = 18 take natural log each side ln[5(2X)] = ln(18) now, as a logarithmic operation the exponent can come down in front of the log sign 2X ln(5) = ln(18) divide thus 2X = ln(18)/ln(5) [ not ln(18/5)!!!! ] 2X = 1.795888947 divide by 2 X = 0.8979444735 -------------------------------check in original equation 5(2*0.8979444735) = 18 18 = 18 -------------checks
What is the answer to 2 minus Ln times 3 minus x equal 0?
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
How do you solve 3e to the power 2x -1 equals 8?
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
Solve or x 2 ln 9 plus 2 ln 5 equals 2 ln x minus 3?
2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)
What is the derivative of ln 1-2x?
3
What is the solution set of 2.5 to the power of 2x equals 8?
2.52x = 8 Taking (natural) logs: 2x*ln(2.5) = ln(8) so 2x = ln(8)/ln(2.5) = 2.0794/0.9163 = 2.2694 approx. and therefore, x = 2.2694/2 = 1.1347 approx.
Integral of ln2x?
I will denote an integral as \int (LaTeX). We can let u = 2x and du = 2dx, and substitute \int ln (2x) dx = (1/2) \int ln u du. Either using integration by parts or by memorization, this is equal to (1/2) u ln u - u + C = (1/2)(2x ln (2x) - 2x) + C, where C is an arbitrary constant.
Why we cancel e by ln?
The definition of the natural log ln of a number is the power that you have to raise e to in order to get that number. Therefore, ln(2x+3) is the power you have to raise e to to get 2x + 3.
What is the value of x given the equation 5 to the power of 1- 2 x equals 0.25 correct to 3 decimal places Please show working out?
51-2x = 0.25 51 * 5-2x = 0.25 5-2x = 0.05 -2x*ln(5) = ln(0.05) x = ln(0.05)/[-2*ln(5)] = 0.931
