Solve Complex Fraction: 2x/(log₈9) ÷ log₉8 Simplification

Question

$\frac{\frac{2x}{\log_89}}{\log_98}=$

00:00 Solve

00:04 Let's break down the multiplication from the fraction

00:14 We'll use the formula of dividing 1 by log

00:17 We'll get the log of the inverse number and base

00:27 We'll use this formula in our exercise

00:40 Let's reduce what we can

00:44 And this is the solution to the question

To solve this problem, we will simplify the expression $\frac{\frac{2x}{\log_8 9}}{\log_9 8}$ .

Step 1: Apply the inverse log property.

The property $\log_b a \times \log_a b = 1$ states that these logs are multiplicative inverses.

Thus, $\log_8 9 \times \log_9 8 = 1$ , meaning $\frac{1}{\log_9 8} = \log_8 9$ .

Step 2: Substitute $\log_8 9$ with $\frac{1}{\log_9 8}$ in the original fraction.

Given the expression is $\frac{\frac{2x}{\log_8 9}}{\log_9 8}$ , it becomes:

$\frac{2x}{\frac{1}{\log_9 8}} \times \frac{1}{\log_9 8} = 2x \times 1 = 2x$ .

Step 3: Simplify the expression.

The multiplication results in the cancelling of the logarithmic terms through the multiplicative inverse relationship.

Therefore, the solution to the problem is $2x$ .

$2x$