How do you solve the equation 4 log12(2) + log12(x) + log12(96)?

Solving the Equation: 4 log₁₂(2) + log₁₂(x) + log₁₂(96)

To solve the equation 4 log₁₂(2) + log₁₂(x) + log₁₂(96) = 0, we will follow a step-by-step approach.

First, let’s break down the individual logarithmic components:

Using the property of logarithms that states k log_b(a) = log_b(a^k), we can write:

4 log₁₂(2) = log₁₂(2⁴) = log₁₂(16)

Since 96 can be factored as 96 = 16 * 6, using properties of logarithms, we can express this as:

log₁₂(96) = log₁₂(16) + log₁₂(6)

Now, we substitute the simplifications from above back into the original equation:

log₁₂(16) + log₁₂(x) + log₁₂(16) + log₁₂(6) = 0

This simplifies to:

2 log₁₂(16) + log₁₂(x) + log₁₂(6) = 0

Now, let’s further simplify it:

log₁₂(x) = -2 log₁₂(16) - log₁₂(6)

Using the property of logarithms that combines logs, we can rewrite this equation:

log₁₂(x) = log₁₂(16^-2) + log₁₂((6)^-1) = log₁₂(16^-2 / 6)

Now, we can exponentiate each side to remove the logarithm:

x = 16^-2 / 6

This simplifies to:

x = 1 / (96) = rac{1}{96}

Thus, the solution to the equation 4 log₁₂(2) + log₁₂(x) + log₁₂(96) = 0 is:

x = rac{1}{96}