Exercise: G.M.L.T. 1

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Exercise: G.M.L.T.  1 < (e^(x)-1)/(ln(x+1)) < (x+1)e^(x)

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Define Domain: First, we need to address the domain of the function (e^(x)-1)/(ln(x+1)) to avoid any undefined expressions. The natural logarithm ln(x+1) is defined for x > -1. Additionally, since we have an inequality involving a fraction, we must ensure that the denominator is not zero, so x cannot be equal to 0. Thus, the domain of x is x > -1 and x ≠ 0.Solve Left Inequality: Next, we will solve the left part of the inequality: 1 < (e^(x)-1)/(ln(x+1)). We will multiply both sides by ln(x+1) to get rid of the denominator, but we must remember that the direction of the inequality will depend on the sign of ln(x+1). Since ln(x+1) is positive for x > 0 and negative for -1 < x < 0, we will have to consider these two cases separately.Consider Cases: For x > 0, ln(x+1) is positive. Multiplying both sides of the inequality by ln(x+1) gives us ln(x+1) < e^(x)-1. We then add 1 to both sides to isolate e^(x) on one side: ln(x+1) + 1 < e^(x).Solve Right Inequality: For -1 < x < 0, ln(x+1) is negative. Multiplying both sides of the inequality by ln(x+1) and reversing the inequality sign gives us ln(x+1) > e^(x)-1. We then add 1 to both sides to isolate e^(x) on one side: ln(x+1) + 1 > e^(x). However, this case leads to a contradiction since e^(x) is always greater than ln(x+1) + 1 for -1 < x < 0. Therefore, there are no solutions in this interval.Simplify Inequality: Now we will solve the right part of the inequality: (e^(x)-1)/(ln(x+1)) < (x+1)e^(x). We will multiply both sides by ln(x+1) to get rid of the denominator. Since we are considering x > 0, ln(x+1) is positive, and the direction of the inequality remains the same: e^(x)-1 < (x+1)e^(x)ln(x+1).Isolate Term: We simplify the inequality by distributing e^(x) on the right side: e^(x)-1 < e^(x)(x+1)ln(x+1) - e^(x)ln(x+1). This simplifies to -1 < e^(x)xln(x+1) - e^(x)ln(x+1).Factor Out: We can then add e^(x)ln(x+1) to both sides to isolate the term with x on one side: e^(x)ln(x+1) - 1 < e^(x)xln(x+1).Divide and Solve: Now we can factor out e^(x)ln(x+1) on the right side: e^(x)ln(x+1) - 1 < e^(x)ln(x+1)(x).Complex Inequality: We divide both sides by e^(x)ln(x+1) to solve for x. Since e^(x)ln(x+1) is positive for x > 0, the direction of the inequality remains the same: 1/e^(x)ln(x+1) - 1/(e^(x)ln(x+1))^2 < x.Summary: This inequality is quite complex to solve algebraically, and it may require numerical methods or graphing techniques to find the exact range of x that satisfies it. However, we can see that as x increases, the left side approaches 0, and thus for sufficiently large x, the inequality will be satisfied.To summarize, we have found that for x > 0, the inequality ln(x+1) + 1 < e^(x) must be satisfied, and for sufficiently large x, the inequality 1/e^(x)ln(x+1) - 1/(e^(x)ln(x+1))^2 < x is satisfied. Therefore, the range of x that satisfies the original inequality is a subset of x > 0, excluding any values that do not satisfy both parts of the inequality.

Exercise: G.M.L.T. $\newline$ 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}

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Exercise: G.M.L.T.\newline 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}

Exercise: G.M.L.T. $\newline$ 1<\frac{e^{x}-1}{\ln (x+1)}<(x+1) e^{x}