1 < (e^(x)-1)/(ln(x-1)) < (x+1)=x

Isolate exponential expression: First, let's address the inequality 1 < (e^(x)-1)/(ln(x-1)). We need to isolate the exponential expression on one side.Multiply by ln(x-1): Multiply both sides of the inequality by ln(x-1) to get rid of the denominator, assuming ln(x-1) is positive (which it must be for the logarithm to be defined and for the inequality to maintain its direction). 1 * ln(x-1) < e^(x) - 1Add 1 to isolate e^(x): Add 1 to both sides of the inequality to isolate e^(x). ln(x-1) + 1 < e^(x)Address second part of inequality: Now, let's address the second part of the inequality (e^(x)-1)/(ln(x-1)) < (x+1)=x. This seems to be a typo or a mistake because (x+1)=x is not an inequality and does not make sense in this context. We will assume that the intended inequality is (e^(x)-1)/(ln(x-1)) < x+1.

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Algebra 1

Solve advanced linear inequalities

Full solution

1<\frac{e^{x}-1}{\ln (x-1)}<(x+1)=x

Isolate exponential expression: First, let's address the inequality 1 < \frac{e^{x}-1}{\ln(x-1)}. We need to isolate the exponential expression on one side.
Multiply by $\ln(x-1)$ : Multiply both sides of the inequality by $\ln(x-1)$ to get rid of the denominator, assuming $\ln(x-1)$ is positive (which it must be for the logarithm to be defined and for the inequality to maintain its direction). $\newline$ 1 \cdot \ln(x-1) < e^{x} - 1
Add $1$ to isolate $e^{x}$ : Add $1$ to both sides of the inequality to isolate $e^{x}$ . $\newline$ \ln(x-1) + 1 < e^{x}
Address second part of inequality: Now, let's address the second part of the inequality (e^{x}-1)/(\ln(x-1)) < (x+1)=x. This seems to be a typo or a mistake because $(x+1)=x$ is not an inequality and does not make sense in this context. We will assume that the intended inequality is (e^{x}-1)/(\ln(x-1)) < x+1.