a. Log8x+log(x-20)=3

Answer 1

To solve the equation log8x + log(x-20) = 3, we can use the properties of logarithms.

First, we can combine the two logarithms using the product rule: log(a) + log(b) = log(a * b).

So, applying this property to the equation, we have log[(8x) * (x-20)] = 3.

Next, using the definition of logarithms, we can rewrite this equation as an exponential equation: 10^3 = (8x) * (x-20).

Simplifying the exponent, we get 1000 = (8x) * (x-20).

Now, we can solve this quadratic equation by setting it equal to zero and factoring or using the quadratic formula.

Setting 1000 - (8x) * (x-20) = 0, we can factor 8x(x-20) - 1000 = 0.

Factoring further, we have 8x(x-20) - 1000 = 0.

Expanding the equation, we get 8x^2 - 160x - 1000 = 0.

Now, we can either factor this quadratic equation or use the quadratic formula to solve for x.

Answer 2

why didn't you finish it ??

8x^2 - 160x - 1000 = 0
x^2 - 20x - 125 = 0
(x-25)(x+5) = 0
x = 25 or x = -5

but in the original equation, x > 0

so x = 25

Answer 3

Apologies for not completing the solution earlier. You are correct in factoring the quadratic equation as:

8x^2 - 160x - 1000 = 0
x^2 - 20x - 125 = 0
(x-25)(x+5) = 0

From here, we obtain two possible solutions: x = 25 or x = -5.

However, when considering the original equation log8x + log(x-20) = 3, we can deduce that x must be greater than 20 (due to the presence of log(x-20)). Additionally, x must be greater than 0 since logarithms are not defined for negative numbers or zero.

Therefore, we discard x = -5 as a valid solution, leaving us with x = 25 as the final solution to the equation log8x + log(x-20) = 3.