Solve the equation. Write the solution set with the exact solutions.

[tex]\[ \log_2 t + \log_2 (t+3) = 2 \][/tex]

There is no solution: [tex]\(\{\}\)[/tex].

The exact solution set is [tex]\(\square\)[/tex].



Answer :

To solve the equation [tex]\(\log_2 t + \log_2 (t + 3) = 2\)[/tex], we can follow these steps:

1. Combine the logarithms:
Using the property of logarithms [tex]\(\log_b (a) + \log_b (c) = \log_b (a \cdot c)\)[/tex], we can rewrite the given equation:
[tex]\[ \log_2 t + \log_2 (t + 3) = \log_2 [t(t + 3)] \][/tex]
Therefore, the equation becomes:
[tex]\[ \log_2 [t(t + 3)] = 2 \][/tex]

2. Rewrite the equation in exponential form:
Since [tex]\(\log_2 [t(t + 3)] = 2\)[/tex], we can write:
[tex]\[ t(t + 3) = 2^2 \][/tex]
Thus:
[tex]\[ t(t + 3) = 4 \][/tex]

3. Form a quadratic equation:
To solve this, expand and rearrange:
[tex]\[ t^2 + 3t - 4 = 0 \][/tex]

4. Solve the quadratic equation:
The quadratic formula [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] can be used, where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -4\)[/tex].

- Calculate the discriminant:
[tex]\[ \text{discriminant} = b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25 \][/tex]

- Find the two solutions:
[tex]\[ t = \frac{-3 \pm \sqrt{25}}{2 \times 1} = \frac{-3 \pm 5}{2} \][/tex]

Therefore:
[tex]\[ t_1 = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \][/tex]
[tex]\[ t_2 = \frac{-3 - 5}{2} = \frac{-8}{2} = -4 \][/tex]

5. Check the solutions:
Since [tex]\(t\)[/tex] must be a positive number for the logarithms to be defined, we discard [tex]\(t = -4\)[/tex] because the logarithm of a negative number is not defined.

Hence, the only valid solution is [tex]\(t = 1\)[/tex].

Solution set:
[tex]\[ \{1\} \][/tex]

So, the exact solution set for the equation [tex]\(\log_2 t + \log_2 (t + 3) = 2\)[/tex] is [tex]\(\{1\}\)[/tex].