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].
