Question

Log3(x-5)+log3(2x-3)=1

Answer 1

Решение на фото .........

Answer 2

To solve the logarithmic equation $\log_3(x-5) + \log_3(2x-3) = 1$, we'll use the properties of logarithms to simplify and then solve for $x$. The key property we'll use here is the product rule of logarithms, which states that $\log_b(a) + \log_b(c) = \log_b(a \cdot c)$.

Step 1: Combine the logarithms using the product rule: $\log_3((x-5)(2x-3)) = 1$

Step 2: Convert the equation to exponential form. In general, if $\log_b(y) = x$, then $b^x = y$. $3^1 = (x-5)(2x-3)$

Step 3: Simplify and solve for $x$: $3 = 2x^2 - 13x + 15$

Step 4: Rearrange the equation to set it equal to zero: $2x^2 - 13x + 12 = 0$

Step 5: Factor the quadratic equation: $(2x - 3)(x - 4) = 0$

Step 6: Set each factor to zero and solve for $x$: $2x - 3 = 0$ or $x - 4 = 0$

Solving for $x$ in each case: $2x = 3$ $\Rightarrow$ $x = \frac{3}{2}$

$x = 4$

Step 7: Check for extraneous solutions. In logarithmic equations, we must ensure that the values of $x$ we found are valid solutions. To be valid, the arguments of the logarithms must be positive.

For $\log_3(x-5) + \log_3(2x-3) = 1$ to hold, $x-5$ and $2x-3$ must both be positive:

For $x = \frac{3}{2}$: $x - 5 = \frac{3}{2} - 5 = \frac{-7}{2}$ (not positive, invalid solution)

For $x = 4$: $x - 5 = 4 - 5 = -1$ (not positive, invalid solution)

So, there are no valid solutions for this equation. Therefore, the original equation $\log_3(x-5) + \log_3(2x-3) = 1$ has no real solutions.