Solve log₄x > 7log₄2: Base-4 Logarithmic Inequality

$7\log_42<\log_4x$

To solve the inequality $7\log_4 2 < \log_4 x$, we will follow these steps:

• Step 1: Simplify $7\log_4 2$ using the logarithm properties.
• Step 2: Write the inequality $\log_4(2^7) < \log_4 x$.
• Step 3: Solve for $x$ by converting the logarithmic inequality into an exponential form.

Let's now proceed with these steps:

Step 1: Using the power property of logarithms, we have $7\log_4 2 = \log_4 (2^7)$. This step simplifies the multiplication into a single logarithmic term.

Step 2: Using substitution in the inequality, we write it as $\log_4 (2^7) < \log_4 x$.

Step 3: Since logarithms are one-to-one functions, we can conclude that if $\log_4 (2^7) < \log_4 x$, then $2^7 < x$. This results from the property $a^c = a^d \iff c = d$ where the bases are equal.

Therefore, the solution to the inequality is $\mathbf{2^7 < x}$.