Solve for x (without using a calculator):
log base 3 of x^2= 2log base 3 of 4-4log base 3 of 5
Solve each of the inequalities:
(a) x^2+2x-3<(or equal to)0
(b) (2x-1)/(3x-2)<(or equal to)1
Any help is appreciated!
log base(3)x^2 = 2log(3)4 - 4log(3)5
= log(3)[4^2/5^4]
Therefore x^2 = 16/625
x = 4/25 = 0.16
<<(a) x^2+2x-3<(or equal to)0 >>
First solve
x^2 +2x -3 =0
(x+3)(x-1) = 0
x = -3 or 1
Between those points, for example at x=0, the function is negative.
(f(0)= -3, for example.)
-3 <_ x <_ 1 is the answer
<<(2x-1)/(3x-2)<(or equal to)1>>
When equal to 1,
2x-1 = 3x-2
x = 1
If you go on the positive side if 1, for example at x = 2,
(2x-1)/(3x-2) = 3/5 which is <1
Any x >_ 1 satisfies the inequality
To solve the equation log base 3 of x^2 = 2log base 3 of 4 - 4log base 3 of 5, we can use the properties of logarithms.
Step 1: Use the power rule to simplify the right side of the equation:
log base 3 of x^2 = log base 3 of 4^2 - log base 3 of 5^4
Step 2: Apply the logarithmic identity log base b of b^a = a:
log base 3 of x^2 = 2 - 4log base 3 of 5
Step 3: Use the multiplication property of logarithms to simplify the right side of the equation:
log base 3 of x^2 = 2 - log base 3 of (5^4)
Step 4: Simplify the right side of the equation further:
log base 3 of x^2 = 2 - log base 3 of 625
Step 5: Use the logarithmic identity log base b of b^a = a:
log base 3 of x^2 = 2 - 4
Step 6: Simplify the equation:
log base 3 of x^2 = -2
Step 7: Rewrite the equation in exponential form:
x^2 = 3^(-2)
Step 8: Simplify the right side of the equation:
x^2 = 1/3^2
Step 9: Simplify further:
x^2 = 1/9
Step 10: Take the square root of both sides to solve for x:
x = ±√(1/9)
Step 11: Simplify the right side of the equation:
x = ±1/3
Therefore, the solution is x = 1/3 or x = -1/3.
To solve inequality (a) x^2 + 2x - 3 ≤ 0, we can find the critical points and use a number line to determine the intervals where the inequality is satisfied.
Step 1: Factor the quadratic equation:
(x + 3)(x - 1) ≤ 0
Step 2: Set each factor equal to zero to find the critical points:
x + 3 = 0 --> x = -3
x - 1 = 0 --> x = 1
Step 3: Create a number line and plot the critical points:
_____________ -3__________ 1 __________
Step 4: Test a value from each interval created to determine whether the inequality is satisfied. Choose a value between -∞ and -3 (e.g., x = -4), a value between -3 and 1 (e.g., x = 0), and a value greater than 1 (e.g., x = 2).
Step 5: Plug each test value into the inequality:
For x = -4: (-4 + 3)(-4 - 1) ≤ 0 --> (-1)(-5) ≤ 0 --> 5 ≤ 0 (False)
For x = 0: (0 + 3)(0 - 1) ≤ 0 --> (3)(-1) ≤ 0 --> -3 ≤ 0 (True)
For x = 2: (2 + 3)(2 - 1) ≤ 0 --> (5)(1) ≤ 0 --> 5 ≤ 0 (False)
Step 6: Determine the intervals where the inequality holds true based on the test values:
(-∞, -3] ∪ (0, 1]
Therefore, the solution to inequality (a) is x ∈ (-∞, -3] ∪ (0, 1].
To solve inequality (b) (2x - 1)/(3x - 2) ≤ 1, we can find the critical points and use a number line to determine the intervals where the inequality is satisfied.
Step 1: Create a number line and plot the critical points:
_____________ 2/3_________ 2 __________
Step 2: Set the numerator and the denominator equal to zero to find the critical points:
2x - 1 = 0 --> x = 1/2
3x - 2 = 0 --> x = 2/3
Step 3: Test a value from each interval created to determine whether the inequality is satisfied. Choose a value less than 2/3 (e.g., x = 0), a value between 2/3 and 1/2 (e.g. x = 1), and a value greater than 1/2 (e.g., x = 2).
Step 4: Plug each test value into the inequality:
For x = 0: (2(0) - 1)/(3(0) - 2) ≤ 1 --> (-1)/(-2) ≤ 1 --> 1/2 ≤ 1 (True)
For x = 1: (2(1) - 1)/(3(1) - 2) ≤ 1 --> (1)/(-1) ≤ 1 --> -1 ≤ 1 (True)
For x = 2: (2(2) - 1)/(3(2) - 2) ≤ 1 --> (3)/(4) ≤ 1 --> 3/4 ≤ 1 (True)
Step 5: Determine the intervals where the inequality holds true based on the test values:
(-∞, 2/3] ∪ (2/3, 1/2)
Therefore, the solution to inequality (b) is x ∈ (-∞, 2/3] ∪ (2/3, 1/2).