Certainly! Let's solve the given logarithmic equation step-by-step.
Given:
[tex]\[
\log_5\left(x^3 - 3x + 2\right) = 2 + \log_5(x - 1)
\][/tex]
First, recall that [tex]\( \log_b(a) + \log_b(c) = \log_b(ac) \)[/tex]. Also, we use [tex]\( \log_b(c) + k = \log_b(b^k \cdot c) \)[/tex].
Rewrite [tex]\( 2 + \log_5(x - 1) \)[/tex]:
[tex]\[
2 + \log_5(x - 1) = \log_5(5^2) + \log_5(x - 1) = \log_5(25) + \log_5(x - 1)
\][/tex]
Using the product rule for logarithms:
[tex]\[
\log_5(25) + \log_5(x - 1) = \log_5(25 \cdot (x - 1)) = \log_5(25(x - 1))
\][/tex]
So the equation can be rewritten as:
[tex]\[
\log_5(x^3 - 3x + 2) = \log_5(25(x - 1))
\][/tex]
Since the logarithms with the same base are equal when their arguments are equal, we get:
[tex]\[
x^3 - 3x + 2 = 25(x - 1)
\][/tex]
Let's simplify this equation:
[tex]\[
x^3 - 3x + 2 = 25x - 25
\][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[
x^3 - 3x + 2 - 25x + 25 = 0
\][/tex]
[tex]\[
x^3 - 28x + 27 = 0
\][/tex]
Now, solve the cubic equation [tex]\( x^3 - 28x + 27 = 0 \)[/tex]. This can be approached by various algebraic methods or using numerical methods, producing the solutions.
After solving, the solutions are:
[tex]\[
\boxed{-\frac{1}{2} + \frac{\sqrt{109}}{2}, -\frac{\sqrt{109}}{2} - \frac{1}{2}}
\][/tex]
So the values of [tex]\( x \)[/tex] that satisfy the given logarithmic equation are:
[tex]\[
x = -\frac{1}{2} + \frac{\sqrt{109}}{2} \quad \text{and} \quad x = -\frac{\sqrt{109}}{2} - \frac{1}{2}
\][/tex]
