The value of `(0.16)^("log"_(2.5)((1)/(3) + (1)/(3^(2)) + (1)/(3^(3)) + …."to" oo))`, is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the expression \( (0.16)^{\log_{2.5}\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots\right)} \), we will follow these steps:

### Step 1: Simplify the Infinite Series
The expression inside the logarithm is an infinite geometric series:
\[ S = \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \]
This is a geometric series where the first term \( a = \frac{1}{3} \) and the common ratio \( r = \frac{1}{3} \).

The sum of an infinite geometric series can be calculated using the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
Substituting the values:
\[ S_{\infty} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 2: Substitute the Sum into the Logarithm
Now we substitute \( S_{\infty} \) back into the logarithm:
\[ (0.16)^{\log_{2.5}\left(\frac{1}{2}\right)} \]

### Step 3: Rewrite \( 0.16 \)
Next, we rewrite \( 0.16 \) as a fraction:
\[ 0.16 = \frac{16}{100} = \frac{4}{25} \]
So we have:
\[ \left(\frac{4}{25}\right)^{\log_{2.5}\left(\frac{1}{2}\right)} \]

### Step 4: Apply Logarithmic Properties
Using the property of logarithms \( a^{\log_b(c)} = c^{\log_b(a)} \), we can rewrite the expression:
\[ \left(\frac{4}{25}\right)^{\log_{2.5}\left(\frac{1}{2}\right)} = \left(\frac{1}{2}\right)^{\log_{2.5}\left(\frac{4}{25}\right)} \]

### Step 5: Simplify \( \frac{4}{25} \)
Now, we can express \( \frac{4}{25} \) as:
\[ \frac{4}{25} = \left(\frac{2}{5}\right)^2 \]
Thus, we can write:
\[ \log_{2.5}\left(\frac{4}{25}\right) = \log_{2.5}\left(\left(\frac{2}{5}\right)^2\right) = 2 \log_{2.5}\left(\frac{2}{5}\right) \]

### Step 6: Substitute Back
Substituting this back gives:
\[ \left(\frac{1}{2}\right)^{2 \log_{2.5}\left(\frac{2}{5}\right)} = \left(\frac{1}{2}\right)^{\log_{2.5}\left(\frac{2}{5}\right)^2} \]

### Step 7: Final Simplification
Using the property of logarithms again, we can simplify:
\[ \left(\frac{1}{2}\right)^{\log_{2.5}\left(\frac{4}{25}\right)} = \left(\frac{4}{25}\right)^{\log_{2.5}\left(\frac{1}{2}\right)} \]
This evaluates to:
\[ \left(\frac{1}{2}\right)^{-2} = 4 \]

### Final Result
Thus, the value of the original expression is:
\[ \boxed{4} \]