Solve for x Provide the exact solutlon_ [4,4] a) 5*-9 b) logkx + 3) + logkx ~ 2) - log(3x + 2)...

Determine whether each $x$ -value is a solution (or an approximate solution) of the equation.
$\ln (2 x+3)=5.8$
(a) $x=\frac{1}{2}(-3+\ln 5.8)$
(b) $x=\frac{1}{2}\left(-3+e^{5.8}\right)$
(c) $x \approx 163.650$

In this problem off exponentially and lower question. We have given that Ellen two X plus three this is two X plus three is equals toe 5.8. And now we have to determine whether each value off Expedia approximate or exact solution off the equation. So we have given the first value off X is here. X equals toe, half multiplied with minus three plus Ln 5.8. This is Ellen five pointed. And now the second value is here. X equals to the stage half minus three. This is minus three plus E to the power 5.8. And now the third part is C. So this is X is approximately equals toe 163 point 650 So now solve it. So take here. This is Ellen two X plus three. So simply this can read it. And I was like here l and two X plus three equals toe 5.8. And now? So, no, we can write it as here e to the power. This is simply Ellen. Two X plus three equals to e to the power five pointed. So this would be simply to x plus three equals. So this is e to the power 5.8. So here two x equals to the stage minus three plus e to the power 5.8 that once exit close toe, half multiplied with minus three plus e to the power 5.8. So that one's here. Be satisfying the value for this month's years. This is satisfying. And now this cannot be satisfied because into the power five point it is not a coastal and five pointed, so this cannot be satisfying. And now, for the third part, we have to solve this equation. So here e to the power they said here e to the power here 5.8 this is 5.8 equals to the stage. 333.29 and minus three would be 327 0.29 Now divided with two, which is equals 263.649 This Eddie kowtow approximately equal to this value. So this will be yes. This is satisfying because this is approximately equals 263 points 650 So this is also satisfying this is alternative. And but this is not satisfying. So for the a part, this is no for the be part. This? Yes. And now for the super it is again, Yes.

In this question, we are going to use right triangle trigonometry. The expression we've been given mention tangent tangent of an angle in a right triangle is opposite over adjacent. So we want to find a diagram that has Annex for the opposite side and a five for the adjacent side. That is diagram A. In diagram A. We would set up an equation that the tangent of 38° is opposite over adjacent. And then to solve for X, we would multiply Both sides of the equation by five. And the expression to solve for X is the expression we were given and asked to match.

U N English in is 44 minus nine. Next is it close to 61. Subjecting board said by 40 food. We get minus off nine. Next is a questo 61 minus 44. My solving this we get minus off Nynex is a questo 17. Divide walked side by minus nine. We get excess a questo minus 17 divided by nine hens. X is it questo minus 17 divided by nine.

This problem has X over two rather than X, but we can ignore that at the beginning and just deal with apathy. End. So what we're doing is we're looking for the angle X over to that has assigned value of Negative Square 3/2, and we know that sign is negative in Quadrant three and quarter four. So let's picture some reference triangles. If sign was negative square three over to, we would have square with three on the opposite to on the high pot news. And we realized we have a 30 60 90 triangle with a 60 degree reference angle. Same thing in quadrant four. So if we were answering and degrees are answers would be 240 degrees on 300 degrees. Put those into radiance and we have four pi over three radiance, and we have five pi over three radiance. And then for generalizing this to all possible answers, we're just gonna add two n pi to both of those. So these are the values of X over to not the values of X to find the values of X. Let's just multiply everything by two. So that gives us X equals eight pi over three plus four pi end and 10 pi over three plus four pi n or for N pi