Final answer:
To solve for y in terms of x in the equation loge(y-4) + 5loge(-2x-4) = 2, we combine the logarithms into a single log, exponentiate both sides, and solve for y to find y = e^2/(-2x-4)^5 + 4.
Step-by-step explanation:
The equation provided is loge(y-4) + 5loge(-2x-4) = 2. To solve for y in terms of x, we follow a series of algebraic and logarithmic properties. First, using the property of logarithms that states the logarithm of a product is the sum of the logarithms, we can rewrite 5loge(-2x-4) as loge((-2x-4)^5). This simplifies our equation to loge(y-4) + loge((-2x-4)^5) = 2, which can then be combined into a single logarithm using the rule that states the sum of two logarithms is the logarithm of the product: loge((y-4)*(-2x-4)^5) = 2.
To remove the logarithm, we exponentiate both sides of the equation using the base of the natural logarithm, e, since loge is the natural logarithm. This gives us (y-4)*(-2x-4)^5 = e^2. We then solve for y to get y = e^2/(-2x-4)^5 + 4.