help solve 8=2^5x * 4^x^2
logarithms
eight equals 2 to the 5x times 4 to the x squared in case I didn't write it correctly.
see the other post
1. The Safety Director of Honda USA took samples at random from the file of minor accidents and classified them according to the time the accident took place. The sample data is as follows:
8am to 9am 6 accidents
9am to 10 am 6 accidents
10am to 11 am 20 accidents
11am to noon 8 accidents
Noon to 1pm 7 accidents
1pm to 2pm 8 accidents
2pm to 3pm 10 accidents
3pm to 4pm 6 accidents
At the 0.01 level of significance, are the accidents evenly distributed throughout the day?
2. For many years TV executives used the guideline that 30 percent of the audience were watching each of the prime-time networks and 10 percent were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the 0.05 level of significance, can we conclude that the old guideline is still reasonable?
To solve the equation 8 = 2^(5x) * 4^(x^2), we can use the properties of logarithms.
In this case, we can take the logarithm of both sides of the equation to simplify it. Let's take the base-2 logarithm (log₂) of both sides:
log₂(8) = log₂(2^(5x) * 4^(x^2))
Next, we can use the power rule of logarithms, which states that log₂(a^b) = b * log₂(a):
log₂(8) = 5x * log₂(2) + x^2 * log₂(4)
Since log₂(2) is equal to 1 and log₂(4) is equal to 2, we can substitute these values into the equation:
log₂(8) = 5x + 2x^2
Now, we have a quadratic equation. We can rearrange it to form a standard quadratic equation:
2x^2 + 5x - log₂(8) = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 2, b = 5, and c = -log₂(8). Substituting these values:
x = (-(5) ± √((5)^2 - 4(2)(-log₂(8)))) / (2(2))
Now we can simplify and calculate the values of x by using a calculator.