Solve the Mixed Logarithm Equation: log7×ln(x) = ln7×log(x²+8x-8)

$\log7\times\ln x=\ln7\cdot\log(x^2+8x-8)$

?=x

To solve this problem, we'll follow these steps:

• Step 1: Set up the equation based on the given: $\log7 \times \ln x = \ln7 \cdot \log(x^2 + 8x - 8)$.
• Step 2: Utilize logarithmic properties and equate the expressions fully.
• Step 3: Transform and solve the derived quadratic equation.

Now, let's work through each step:

Step 1: Consider the given equation: $\log7 \times \ln x = \ln7 \cdot \log(x^2 + 8x - 8)$.

Step 2: We can leverage the commutative property of multiplication to rewrite the equation:
$\frac{\ln x}{\ln7} = \frac{\log(x^2 + 8x - 8)}{\log7}$.

Cross-multiplying gives:
$\ln x \cdot \log7 = \ln7 \cdot \log(x^2 + 8x - 8)$.

Rule out common denominators to get equality in logs, rewritten equation:
$\ln x = \log(x^2 + 8x - 8)$.

Step 3: Assume the simplest corresponding argument equality:
$x = x^2 + 8x - 8$ (consider logarithmic domain; check/simplify where equal in rational space) then solve for real roots / positively defined solutions:

Rearrange to form a quadratic equation:
$0 = x^2 + 8x - x - 8 = x^2 + 7x - 8$

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 7$, $c = -8$:

$x = \frac{-7 \pm \sqrt{49 + 32}}{2}$

$x = \frac{-7 \pm \sqrt{81}}{2}$

$x = \frac{-7 \pm 9}{2}$

This results in two possible solutions:
$x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-16}{2} = -8$

Since logarithms require positive values:
Available within positive domain: $x = 1$

Therefore, the solution to the problem is $x = 1$.