Use logarithms to solve.
 
e2x−ex−110=0
 
Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value).

Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example, c*ln(h).
 
Hints:

Remember by the rule of exponents, that e2x=(ex)2 so that you can try to factor this as (ex+something)(ex+somethingelse) where the "something" and "something else" can be positive or negative numbers.

Alternately, you could let y=ex and this equation would be y2−y−110=0 and you can solve first for y and then for x.




We are tasked with solving the equation:

\[
e^{2x} - e^x - 110 = 0
\]

### Step 1: Substitution

Let \( y = e^x \). This transforms the equation into a quadratic form:

\[
y^2 - y - 110 = 0
\]

### Step 2: Solving the quadratic equation

Now we solve the quadratic equation \( y^2 - y - 110 = 0 \). Using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = -1 \), and \( c = -110 \). Substituting these values in:

\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-110)}}{2(1)}
\]

\[
y = \frac{1 \pm \sqrt{1 + 440}}{2}
\]

\[
y = \frac{1 \pm \sqrt{441}}{2}
\]

\[
y = \frac{1 \pm 21}{2}
\]

Thus, the two possible solutions for \( y \) are:

\[
y = \frac{1 + 21}{2} = 11 \quad \text{or} \quad y = \frac{1 - 21}{2} = -10
\]

### Step 3: Solve for \( x \)

Recall that \( y = e^x \). Therefore, we have two cases to solve:

1. \( e^x = 11 \)
2. \( e^x = -10 \)

However, since \( e^x \) is always positive for real numbers, the second equation \( e^x = -10 \) has no real solution.

For \( e^x = 11 \), we take the natural logarithm of both sides:

\[
x = \ln(11)
\]

Thus, the exact solution is:

\[
x = \ln(11)
\]

Would you like more details on any step or have any questions?

---

Here are 5 related questions to expand on this concept:

1. How can logarithms be used to solve more complex exponential equations?
2. What is the significance of the quadratic formula in solving equations like this one?
3. How do we know that \( e^x \) must be positive, and how does that limit the solution?
4. What other methods besides substitution could be used to solve such equations?
5. Can the quadratic formula be applied to non-exponential equations?

**Tip**: Remember that the logarithmic function is the inverse of the exponential function, so taking the natural logarithm helps solve equations involving \( e^x \).

Use logarithms to solve. e^(2x) − e^x − 110 = 0 Enter the exact answer (i.e. keep your answer in exponential or logarithmic form, you do not need to calculate its numeric value).

Learn how to solve the exponential equation e^(2x) − e^x − 110 = 0 using logarithms and quadratic formulas. This problem highlights key concepts in solving exponential equations with logarithmic solutions. Suitable for students in Grades 10-12.

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