Certainly! Let's solve the given equation step by step.
The given equation is:
[tex]\[
(15 - 5x) - \log(3x - 2) - 2 = 0
\][/tex]
### Step 1: Simplify the Equation
First, we simplify the expression within the equation:
[tex]\[
15 - 5x - \log(3x - 2) - 2 = 0
\][/tex]
Combine constants:
[tex]\[
(15 - 2) - 5x - \log(3x - 2) = 0
\][/tex]
[tex]\[
13 - 5x - \log(3x - 2) = 0
\][/tex]
### Step 2: Isolate the Logarithmic Function
Next, move the logarithmic term to one side of the equation:
[tex]\[
13 - 5x = \log(3x - 2)
\][/tex]
### Step 3: Exponentiate Both Sides
To eliminate the logarithmic function, we exponentiate both sides with base [tex]\(e\)[/tex]:
[tex]\[
e^{13 - 5x} = 3x - 2
\][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Now our equation is:
[tex]\[
e^{13 - 5x} = 3x - 2
\][/tex]
This is a transcendental equation involving [tex]\(x\)[/tex] in both the exponent and base of different terms. Such equations typically do not have a straightforward algebraic solution.
### Step 5: Use the Lambert W Function
The Lambert W function, denoted as [tex]\(W(z)\)[/tex], satisfies [tex]\(Z = W(Z)e^{W(Z)}\)[/tex]. We can manipulate the equation to apply this function.
Rewriting:
[tex]\[
e^{13} \cdot e^{-5x} = 3x - 2
\][/tex]
Let’s set:
[tex]\[
u = -5x
\][/tex]
Substituting back into the equation:
[tex]\[
e^{13} \cdot e^{u} = 3\left(\frac{-u}{5}\right) - 2
\][/tex]
Simplifying further:
[tex]\[
e^{13} \cdot e^{u} = -\frac{3u}{5} - 2
\][/tex]
After isolating [tex]\( e^u \)[/tex]:
[tex]\[
e^{13} \cdot e^{u} + 2 = -\frac{3u}{5}
\][/tex]
Multiply through by 5:
[tex]\[
5e^{13}e^u + 10 = -3u
\][/tex]
Now, to isolate [tex]\(e^{u}\)[/tex], we have:
[tex]\[
5e^{u + 13} = -3u - 10
\][/tex]
Express [tex]\(u\)[/tex] as:
[tex]\[
u = \frac{-10 - 3W \left( \frac{5e^{13}}{-3} \right) }{-3}
\][/tex]
Thus:
[tex]\[
x = \frac{2}{3} + \frac{1}{5} W\left( \frac{5e^{29/3}}{3} \right)
\][/tex]
### Conclusion
Hence, the solution to the equation [tex]\( (15 - 5x) - \log(3x - 2) - 2 = 0 \)[/tex] is given by:
[tex]\[
x = \frac{2}{3} + \frac{1}{5}W\left(\frac{5e^{29/3}}{3}\right)
\][/tex]
This solution involves the Lambert W function, which is typically evaluated using numerical methods or special function libraries in mathematical software.
