x triples every second. How will `log_(2)x` change every second

Options

A. If will double every second
B. It will triple every second
C. It increases by cosntant amount every second
D. None of these

Answer

To solve the problem of how `log_2(x)` changes every second when `x` triples every second, we can break it down step by step.

### Step 1: Understand the initial condition
Let’s denote the initial value of `x` at time `t = 0` as `x_0`. Therefore, we have:
\[ x_0 = x \]

### Step 2: Determine the value of `x` after 1 second
Since `x` triples every second, after 1 second, the value of `x` becomes:
\[ x_1 = 3x_0 \]

### Step 3: Calculate `log_2(x)` at time `t = 0`
Now, we calculate `log_2(x)` at time `t = 0`:
\[ \log_2(x_0) = \log_2(x) \]

### Step 4: Calculate `log_2(x)` at time `t = 1`
Next, we calculate `log_2(x)` after 1 second:
\[ \log_2(x_1) = \log_2(3x_0) \]

### Step 5: Use the logarithmic property
Using the property of logarithms that states `log_b(mn) = log_b(m) + log_b(n)`, we can rewrite the expression:
\[ \log_2(3x_0) = \log_2(3) + \log_2(x_0) \]

### Step 6: Substitute the initial value
Now, substituting `log_2(x_0)` back into the equation:
\[ \log_2(3x_0) = \log_2(3) + \log_2(x) \]

### Step 7: Conclusion about the change in `log_2(x)`
From the above steps, we see that every second, when `x` triples, `log_2(x)` increases by a constant amount, which is `log_2(3)`. Therefore, after each second, we can express the change in `log_2(x)` as:
\[ \text{Change in } \log_2(x) = \log_2(3) \]

### Summary
- At `t = 0`, `log_2(x) = log_2(x)`.
- At `t = 1`, `log_2(x) = log_2(3) + log_2(x)`.
- Thus, `log_2(x)` increases by `log_2(3)` every second.