Taking square roots:

−
1
≤
ln
⁡
x
≤
1
−1≤lnx≤1

Converting to exponential form:

e
−
1
≤
x
≤
e
1
e 
−1
 ≤x≤e 
1
 

Which gives:

1
e
≤
x
≤
e
e
1
​
 ≤x≤e what did you do in this part?

Let's break down what happens in that part of the solution more clearly:

### Original Step:
We started with the inequality after squaring:

\[
1 \geq (\ln x)^2
\]

This can be rewritten as:

\[
-1 \leq \ln x \leq 1
\]

### Converting to Exponential Form:
The logarithm function, \(\ln x\), represents the exponent to which the base \(e\) must be raised to obtain \(x\). 

So, if \(\ln x \leq 1\), it means:

\[
x \leq e^1 = e
\]

And if \(\ln x \geq -1\), it means:

\[
x \geq e^{-1} = \frac{1}{e}
\]

### Combining the Results:
Putting these two together, we get:

\[
\frac{1}{e} \leq x \leq e
\]

This interval tells us that \(x\) must be between \(\frac{1}{e}\) (which is approximately 0.3679) and \(e\) (which is approximately 2.718) for the original inequality \(1 - (\ln x)^2 \geq 0\) to hold true.

So, the steps were:

1. **Square Root Inequality:** Solving \(1 \geq (\ln x)^2\) led to \(-1 \leq \ln x \leq 1\).
2. **Exponential Conversion:** Converting \(\ln x \leq 1\) gives \(x \leq e\), and converting \(\ln x \geq -1\) gives \(x \geq \frac{1}{e}\).
3. **Interval Combination:** The valid interval is then \(\frac{1}{e} \leq x \leq e\).

Would you like to explore any further details, or do you have any other questions?

Here are 5 related questions you might consider:

1. Why is the interval \(-1 \leq \ln x \leq 1\) critical for solving the inequality?
2. How does the exponential function impact the solution interval?
3. What happens to the solution if we deal with a different logarithm base?
4. How can the relationship between logarithms and exponents help in solving similar inequalities?
5. Can you solve the same problem using a graph of the function?

**Tip:** When working with inequalities involving logarithms, converting between logarithmic and exponential forms can provide valuable insights into the range of possible solutions.

Explore how to solve logarithmic inequalities using exponential conversions. Learn step-by-step with detailed explanations and examples. Suitable for advanced high school students and anyone interested in logarithmic and exponential functions.

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