Question

F(x) = log5 (2x – 1)

Answer 1

Ответ:

$f(x) = log_{5}(2x - 1) \\ log_{5}(2x - 1) = 0 \: ; \: \: \: x > \frac{1}{2} \\ 2x - 1 = 1 \\ 2x = 1 + 1 \\ 2x = 2 \\ x = 1 \: \: \: \: ; \: \: \: x > \frac{1}{2}$

Answer 2

The function $f(x) = \log_5(2x - 1)$ is a logarithmic function with base 5. Let me explain how this function works:

1. Domain: The domain of this function is all real numbers greater than $\frac{1}{2}$ because the expression inside the logarithm, $2x - 1$, must be greater than zero. In other words, $2x - 1 > 0$, which implies $2x > 1$, and $x > \frac{1}{2}$.

2. Range: The range of this function is all real numbers. Logarithmic functions can produce any real number as their output.

3. Graph: The graph of this logarithmic function will look like the typical logarithmic curve. It will start at the point $(\frac{1}{2}, 0)$ and then rise slowly as $x$ increases. The curve will approach but never touch the x-axis as $x$ approaches infinity. It will also never go below the x-axis because the domain does not allow it to take negative values.

4. Asymptotes: The vertical asymptote of this function occurs at $x = \frac{1}{2}$ because that's where the denominator ($2x - 1$) becomes zero. There are no horizontal asymptotes.

5. Special points:

   • At $x = \frac{1}{2}$, the function is undefined because $\log_5(0)$ is undefined.
   • At $x = 1$, the function becomes $\log_5(2 - 1) = \log_5(1) = 0$, so the graph passes through the point $(1, 0)$.

Overall, the graph of $f(x) = \log_5(2x - 1)$ is a logarithmic curve that rises slowly as $x$ increases, with a vertical asymptote at $x = \frac{1}{2}$ and passing through the point $(1, 0)$.