Question

Solve the logarithmic equation. Select all correct solutions.

log(5x^2−7x)=log(4x^2−5x+3)

Answer 1

To solve the logarithmic equation \( \log(5x^2 - 7x) = \log(4x^2 - 5x + 3) \), we can use the property of logarithms that states if \( \log(a) = \log(b) \), then \( a = b \) (provided \( a > 0 \) and \( b > 0 \)).

Thus, we set:

\[
5x^2 - 7x = 4x^2 - 5x + 3
\]

Now, we rearrange this equation to isolate all terms on one side:

\[
5x^2 - 7x - 4x^2 + 5x - 3 = 0
\]

Combining like terms, we have:

\[
x^2 - 2x - 3 = 0
\]

Next, we can factor the quadratic equation:

\[
(x - 3)(x + 1) = 0
\]

Setting each factor equal to zero gives us the solutions:

\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

Now we need to check if these solutions are valid by ensuring that the arguments of the logarithms are positive:

1. For \( x = 3 \):
\[
5(3^2) - 7(3) = 5(9) - 21 = 45 - 21 = 24 \quad (> 0)
\]
\[
4(3^2) - 5(3) + 3 = 4(9) - 15 + 3 = 36 - 15 + 3 = 24 \quad (> 0)
\]
Since both arguments are positive, \( x = 3 \) is valid.

2. For \( x = -1 \):
\[
5(-1)^2 - 7(-1) = 5(1) + 7 = 5 + 7 = 12 \quad ( > 0)
\]
\[
4(-1)^2 - 5(-1) + 3 = 4(1) + 5 + 3 = 4 + 5 + 3 = 12 \quad ( > 0)
\]
Since both arguments are positive, \( x = -1 \) is also valid.

Thus, the correct solutions to the logarithmic equation are:

\[
\boxed{3} \quad \text{and} \quad \boxed{-1}
\]