To solve the equation $\log_4 2 = 3x - 8$, follow these steps:

Step 1: Evaluate the logarithm

The logarithm $\log_4 2$ asks, "To what power must 4 be raised to get 2?" This can be expressed as:

$\log_4 2 = \frac{1}{2}$

Step 2: Substitute the logarithm value into the equation

Now, substitute $\log_4 2$ with $\frac{1}{2}$ in the equation:

$\frac{1}{2} = 3x - 8$

Step 3: Solve for $x$

Next, solve the equation for $x$:

1. Add 8 to both sides: $\frac{1}{2} + 8 = 3x$

2. Convert 8 to a fraction with a denominator of 2: $\frac{1}{2} + \frac{16}{2} = 3x$

3. Add the fractions: $\frac{17}{2} = 3x$

4. Divide both sides by 3 to solve for $x$: $x = \frac{17}{6}$

So, the solution is:

$x = \frac{17}{6}$

Would you like more details on any step or have any questions?

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Here are 5 related questions to consider:

1. How would you solve the equation $\log_4 8 = 2x - 5$?
2. Can you express $\log_b a$ in terms of common logarithms (base 10)?
3. How would you solve the equation $\log_2 (x+3) = 4$?
4. What is the value of $x$ in $\log_{10} x = 3$?
5. How do you change the base of a logarithm from base $b$ to base $c$?

Tip: When dealing with logarithms, remember that $\log_b a$ can be rewritten as $\frac{\log_c a}{\log_c b}$ for any positive base $c$. This is called the change of base formula.