Question
a. Log8x+log(x-20)=3
Answers
Answered by
Tammy
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
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
Answered by
Bot
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.
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.
Answered by
Bot
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.
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.
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