( x2 + 6x-3) x2-4x = (5-x)x2-4x
32x+3- 6.3x+1- 9 = 0
6log5=a and 5log4=b
4log0,24=…..
Are these 4 separate questions?
Convert the second equation to
25.7 x = 5
then divide both sides by 25.7.
Show your work and we will be glad to critique it for you
Yes, these are four separate questions. I will address each question individually.
Question 1: (x^2 + 6x - 3) * (x^2 - 4x) = (5 - x) * (x^2 - 4x)
To solve this equation, you can expand and combine like terms on both sides of the equation.
First, expand both sides:
x^4 - 4x^3 + 6x^3 - 24x^2 - 3x^2 + 12x = 5x^2 - 20x - x^3 + 4x^2
Next, combine like terms:
x^4 + 2x^3 - 27x^2 + 20x = 9x^2 - 20x
Now, move all terms to one side of the equation to get:
x^4 + 2x^3 - 36x^2 = 0
This is a fourth-degree polynomial equation. To solve it, you can try factoring, using the rational root theorem, or using numerical methods like graphing or a calculator.
Question 2: 32x + 3 - 6.3x + 1 - 9 = 0
To simplify and solve this equation, combine like terms:
32x - 6.3x + 3 + 1 - 9 = 0
25.7x - 5 = 0
To solve for x, move the constant term to the right side:
25.7x = 5
Finally, divide both sides of the equation by 25.7:
x = 5/25.7
Simplifying the right-hand side gives:
x ≈ 0.1945 (rounded to four decimal places)
Question 3: 6log5 = a and 5log4 = b
To solve for a and b:
For the first equation, divide both sides by 6:
log5 = a/6
For the second equation, divide both sides by 5:
log4 = b/5
These equations cannot be further simplified unless you know the value of log5 or log4. If you provide the values, I can compute the answers accordingly.
Question 4: 4log0.24 = ?
To evaluate this logarithmic expression, you can use the properties of logarithms. Specifically, you can rewrite log0.24 as the logarithm of the reciprocal of 0.24 (which is 1/0.24) and then apply the power rule of logarithms.
Start by rewriting the expression:
4log0.24 = 4log(1/0.24)
Next, using the power rule of logarithms, write the expression as:
4(-log0.24)
Finally, you can evaluate -log0.24 using a calculator or a logarithmic table to get the numerical value.