$$ \left. \begin{array} { l l } { 5.3 } & { \frac { e ^ { x } - e ^ { - x } } { e ^ { x } + e ^ { - x } } = \frac { 1 } { 2 e } } \ { 5.4 } & { e ^ { 2 \ln ( 2 x - 3 ) } = 3 x } \end{array} ight. $$

Question

$$ \left. \begin{array} { l l } { 5.3 } & { \frac { e ^ { x } - e ^ { - x } } { e ^ { x } + e ^ { - x } } = \frac { 1 } { 2 e } } \ { 5.4 } & { e ^ { 2 \ln ( 2 x - 3 ) } = 3 x } \end{array} ight. $$

UpStudy AI · Accepted Answer

**5.3**

We need to solve
$$
\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{1}{2e}.
$$
Observe that the left-hand side is the hyperbolic tangent,
$$
\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}},
$$
so the equation becomes
$$
\tanh x = \frac{1}{2e}.
$$

To solve for $$ x $$, we take the inverse hyperbolic tangent:
$$
x = \operatorname{arctanh}\left(\frac{1}{2e}\right).
$$
The formula for the inverse hyperbolic tangent is:
$$
\operatorname{arctanh}(y) = \frac{1}{2}\ln\left(\frac{1+y}{1-y}\right).
$$
Here, $$ y = \frac{1}{2e} $$. Thus,
$$
x = \frac{1}{2}\ln\left(\frac{1+\frac{1}{2e}}{1-\frac{1}{2e}}\right).
$$

---

**5.4**

We start with
$$
e^{2\ln(2x-3)} = 3x.
$$
Using the property $$ e^{\ln a^b} = a^b $$, we rewrite the left-hand side as:
$$
e^{2\ln(2x-3)} = (2x-3)^2.
$$
Thus, the equation becomes:
$$
(2x-3)^2 = 3x.
$$

Expanding the square:
$$
(2x-3)^2 = 4x^2 - 12x + 9,
$$
the equation is:
$$
4x^2 - 12x + 9 = 3x.
$$
Subtract $$ 3x $$ from both sides:
$$
4x^2 - 15x + 9 = 0.
$$

We solve this quadratic equation using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$
with $$ a = 4 $$, $$ b = -15 $$, and $$ c = 9 $$. First, compute the discriminant:
$$
\Delta = (-15)^2 - 4(4)(9) = 225 - 144 = 81.
$$
Then,
$$
x = \frac{15 \pm \sqrt{81}}{8} = \frac{15 \pm 9}{8}.
$$

This gives two potential solutions:
$$
x = \frac{15 + 9}{8} = \frac{24}{8} = 3,
$$
$$
x = \frac{15 - 9}{8} = \frac{6}{8} = \frac{3}{4}.
$$

Since the original equation involves $$\ln(2x-3)$$, the argument $$2x-3$$ must be positive:
$$
2x-3 > 0 \quad \Rightarrow \quad x > \frac{3}{2}.
$$
Only $$ x = 3 $$ satisfies this condition. Therefore, the valid solution is:
$$
x = 3.
$$

---

**Answer Summary:**

- For 5.3: $$\displaystyle x = \frac{1}{2}\ln\left(\frac{1+\frac{1}{2e}}{1-\frac{1}{2e}}\right)$$.
- For 5.4: $$x = 3$$.
**5.3** $$ x = \frac{1}{2}\ln\left(\frac{1+\frac{1}{2e}}{1-\frac{1}{2e}}\right) $$ **5.4** $$ x = 3 $$

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