Solve Mixed Logarithmic Equation: ln(8x) and log₇(e²) Problem

Video Solution

Solution Steps

00:00 Solve

00:03 We'll use the formula to convert to log

00:12 We'll use this formula in our exercise

00:25 We'll use the formula for logical multiplication, we'll get the log of their product

00:33 We'll use this formula in our exercise

00:58 We'll use the formula for logical subtraction, we'll get the log of their quotient

01:08 We'll use this formula in our exercise

01:23 Let's simplify as much as possible

01:33 We'll use the formula for logical multiplication, we'll switch between the numbers

01:48 We'll use this formula in our exercise

02:08 We'll calculate the log, and substitute the solution in the exercise

02:28 We got an equation

02:33 Let's find the domain for finding the solution

02:48 And this is the solution to the question

Step-by-Step Solution

To solve the problem, we proceed as follows:

Given the equation:

$\ln 8x \times \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$

Step 1: Express $\ln 8x$ using the change of base formula:
$\ln 8x = \frac{\log_7 (8x)}{\log_7 e}$
Step 2: Substitute into the original equation:
$\frac{\log_7 (8x)}{\log_7 e} \cdot \log_7 e^2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$
Step 3: Simplify using $\log_7 e^2 = 2 \log_7 e$ :
$\frac{\log_7 (8x)}{\log_7 e} \cdot 2 \log_7 e = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$
Step 4: Cancel $\log_7 e$ and simplify:
$\log_7 (8x) \cdot 2 = 2(\log_7 8 + \log_7 x^2 - \log_7 x)$
Step 5: Cancel 2 on both sides:
$\log_7 (8x) = \log_7 8 + \log_7 x^2 - \log_7 x$
Step 6: Use the properties of logarithms:
$\log_7 (8x) = \log_7 8 + \log_7 \frac{x^2}{x}$
Step 7: Simplify $\log_7 \frac{x^2}{x}$ :
$\log_7 (8x) = \log_7 8 + \log_7 x$
Step 8: Use properties $\log_b m + \log_b n = \log_b (mn)$ :
$\log_7 (8x) = \log_7 (8x)$
Step 9: This equality is true for all $x > 0$ , considering domain restrictions:
$\text{For } x > 0$

Thus, the solution is valid for all $x$ such that $x > 0$

Therefore, the correct solution is, For all $\mathbf{x > 0}$ .