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EA Summer 2023 Algebra II
Evaluating Logarithmic Expressions
Pre-Test

What is the solution of [tex]\log(t-3) = \log(17 - 4t)[/tex]?

A. 4
B. 5
C. 15
D. 20

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Sagot :

To solve the equation [tex]\(\log (t - 3) = \log (17 - 4t)\)[/tex], follow these detailed steps:

1. Understand the property of logarithms:
The equation [tex]\(\log (a) = \log (b)\)[/tex] implies that [tex]\(a = b\)[/tex], provided [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are both positive.

2. Apply the property:
Given [tex]\(\log (t - 3) = \log (17 - 4t)\)[/tex], we can set the arguments of the logarithms equal to each other:
[tex]\[ t - 3 = 17 - 4t \][/tex]

3. Solve for [tex]\(t\)[/tex]:
- Start by isolating [tex]\(t\)[/tex] on one side. Add [tex]\(4t\)[/tex] to both sides to combine the [tex]\(t\)[/tex] terms:
[tex]\[ t + 4t - 3 = 17 \][/tex]
This simplifies to:
[tex]\[ 5t - 3 = 17 \][/tex]

- Next, add 3 to both sides to isolate the term with [tex]\(t\)[/tex]:
[tex]\[ 5t - 3 + 3 = 17 + 3 \][/tex]
This simplifies to:
[tex]\[ 5t = 20 \][/tex]

- Finally, divide by 5 to solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{20}{5} = 4 \][/tex]

4. Check the solution:
For the solution [tex]\(t = 4\)[/tex] to be valid, the arguments of the original logarithms must be positive:
- Check [tex]\(t - 3 > 0\)[/tex]:
[tex]\[ 4 - 3 = 1 > 0 \][/tex]
- Check [tex]\(17 - 4t > 0\)[/tex]:
[tex]\[ 17 - 4 \times 4 = 17 - 16 = 1 > 0 \][/tex]

Both conditions are satisfied, so the solution [tex]\(t = 4\)[/tex] is valid.

Therefore, the solution to the equation [tex]\(\log (t - 3) = \log (17 - 4t)\)[/tex] is [tex]\(\boxed{4}\)[/tex].