8.1 Solve for \( x \) : \[ \log (x+3)+\log x=1 \] 8.2 Show that the above equation will have an irrational root if the base is changed to 2 , and a rational root if the base is changed to 4 .

To solve the equation \( \log (x+3) + \log x = 1 \), we can use the property of logarithms that states \( \log a + \log b = \log(ab) \). Thus, the equation simplifies to: \[ \log (x(x+3)) = 1 \] This means that: \[ x(x+3) = 10^1 = 10 \] Expanding this gives: \[ x^2 + 3x - 10 = 0 \] Now, we can apply the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1, b = 3, c = -10 \): \[ b^2 - 4ac = 3^2 - 4(1)(-10) = 9 + 40 = 49 \] So, the roots are: \[ x = \frac{{-3 \pm 7}}{2} \] Calculating the two possible values gives: \[ x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-10}{2} = -5 \] Since \( x \) must be positive, we have \( x = 2 \) as the solution. Now, let's discuss what happens when we change the base of the logarithm. If the base is changed to 2, we convert our initial equation to: \[ \log_2 ((x+3) x) = 1 \] This simplifies to: \[ (x+3)x = 2^1 = 2 \] The quadratic now is: \[ x^2 + 3x - 2 = 0 \] Using the quadratic formula: \[ b^2 - 4ac = 3^2 - 4(1)(-2) = 9 + 8 = 17 \] Since 17 is not a perfect square, the solutions will be irrational. On changing the base to 4, the equation becomes: \[ \log_4 ((x+3)x) = 1 \] Which simplifies to: \[ (x+3)x = 4^1 = 4 \] The resultant equation is: \[ x^2 + 3x - 4 = 0 \] Applying the quadratic formula here yields: \[ b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25 \] Since 25 is a perfect square, the roots will be rational numbers. Thus, we conclude - when the base is 2, we get an irrational root, while with a base of 4, we find a rational root!