Give the real solution set for each of the following:
42) log(6)(x^2+11) + log(6)x = 1 + log (6)(x^2+1)
this is 11th grade highschool math. if you don't know this drop out now before you waste aany more of your parents' money
To find the real solution set for the given equation:
1. Combine the logarithmic terms on the left-hand side using the properties of logarithms.
log(6)(x^2 + 11) + log(6)x = 1 + log(6)(x^2 + 1)
Apply the product rule of logarithms:
log(6)[x^2(x^2 + 11)] = log(6)(x^2 + 1)
2. Remove the logarithms by taking the exponential of both sides with base 6:
6^[log(6)[x^2(x^2 + 11)]] = 6^[log(6)(x^2 + 1)]
Simplifying the left-hand side:
x^2(x^2 + 11) = x^2 + 1
3. Expand and rearrange the equation:
x^4 + 11x^2 = x^2 + 1
Subtract x^2 + 1 from both sides:
x^4 + 10x^2 - 1 = 0
4. This is a quadratic equation in terms of x^2. Let's substitute y = x^2 to simplify it:
y^2 + 10y - 1 = 0
5. Solve this quadratic equation for y using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Substituting a = 1, b = 10, and c = -1:
y = (-10 ± √(10^2 - 4 * 1 * -1)) / (2 * 1)
Simplifying:
y = (-10 ± √(100 + 4)) / 2
y = (-10 ± √104) / 2
y ≈ (-10 ± 10.198) / 2
6. Solve for y to find the values of x:
For y = (-10 + 10.198) / 2:
y ≈ 0.099
x^2 = 0.099
x ≈ ±0.315
For y = (-10 - 10.198) / 2:
y ≈ -10.099
x^2 = -10.099 (not possible since square roots of negative numbers are not real)
7. Since x^2 cannot be a negative value, the only real solution is:
x ≈ ±0.315
The real solution set for the given equation is approximately x = ±0.315.