Key Concepts and Formulas

• Sum of an Infinite Geometric Progression (GP): The sum of an infinite geometric series with first term $a$ and common ratio $r$ is $S_\infty = \frac{a}{1-r}$, provided $|r| < 1$.
• Logarithm Properties:
  • $b^{\log_b x} = x$ (Identity)
  • $k \log_b x = \log_b (x^k)$ (Power Rule)
• Exponent Properties:
  • $(a^m)^n = a^{mn}$
  • $a^{-n} = \frac{1}{a^n}$
• Number Conversion: Decimals can be converted to fractions (e.g., $0.16 = \frac{16}{100}$, $2.5 = \frac{25}{10}$).

Step-by-Step Solution

We need to evaluate the expression: $E = {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$

Step 1: Evaluate the sum of the infinite geometric series. The argument of the logarithm is an infinite geometric series: $S = {1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....$ The first term is $a = \frac{1}{3}$. The common ratio is $r = \frac{1/3^2}{1/3} = \frac{1}{3}$. Since $|r| = |\frac{1}{3}| < 1$, the series converges. The sum is: $S_\infty = \frac{a}{1-r} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$ Now the expression becomes: $E = {\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$

Step 2: Convert the numbers in the expression to fractional forms. The base of the exponent is $0.16 = \frac{16}{100} = \frac{4}{25}$. The base of the logarithm is $2.5 = \frac{25}{10} = \frac{5}{2}$. The expression is now: $E = {\left( {\frac{4}{25}} \right)^{{{\log }_{5/2}}\left( {{1 \over 2}} \right)}}$

Step 3: Express the base of the exponent in terms of the base of the logarithm. We observe that $\frac{4}{25} = \left(\frac{2}{5}\right)^2$. Also, $\frac{2}{5} = \left(\frac{5}{2}\right)^{-1}$. Therefore, $\frac{4}{25} = \left(\left(\frac{5}{2}\right)^{-1}\right)^2 = \left(\frac{5}{2}\right)^{-2}$. Substituting this into the expression: $E = {\left( {\left(\frac{5}{2}\right)^{-2}} \right)^{{{\log }_{5/2}}\left( {{1 \over 2}} \right)}}$

Step 4: Apply exponent and logarithm properties to simplify. Using the exponent property $(a^m)^n = a^{mn}$: $E = {\left(\frac{5}{2}\right)^{-2 \cdot {{\log }_{5/2}}\left( {{1 \over 2}} \right)}}$ Using the logarithm power rule $k \log_b x = \log_b (x^k)$: $E = {\left(\frac{5}{2}\right)^{{{\log }_{5/2}}\left( {\left( {{1 \over 2}} \right)^{-2}} \right)}}$ Simplify the argument of the logarithm: $\left( {{1 \over 2}} \right)^{-2} = \left(2^{-1}\right)^{-2} = 2^{(-1)(-2)} = 2^2 = 4$ So the expression becomes: $E = {\left(\frac{5}{2}\right)^{{{\log }_{5/2}}\left( {4} \right)}}$

Step 5: Apply the fundamental logarithmic identity. Using the identity $b^{\log_b x} = x$: $E = 4$

Common Mistakes & Tips

• Order of Operations: Always simplify the innermost part of the expression (the infinite series) first.
• Fraction Conversion: Converting decimals to fractions is crucial for identifying relationships between bases and arguments in logarithms and exponents.
• Logarithm Base Matching: Aim to express the base of the exponent in terms of the base of the logarithm to utilize the $b^{\log_b x} = x$ identity.

Summary

The problem requires evaluating an expression containing an infinite geometric series within a logarithm, which is then used as an exponent. By first calculating the sum of the infinite geometric series to be $\frac{1}{2}$, we simplified the expression. Converting all numerical bases and arguments to fractions ($0.16 = \frac{4}{25}$, $2.5 = \frac{5}{2}$) allowed us to express the base of the main exponent in terms of the logarithm's base ($\frac{4}{25} = (\frac{5}{2})^{-2}$). Applying the exponent rule $(a^m)^n = a^{mn}$ and the logarithm power rule $k \log_b x = \log_b (x^k)$ transformed the expression to $(\frac{5}{2})^{\log_{5/2}(4)}$. Finally, the fundamental logarithmic identity $b^{\log_b x} = x$ yielded the result of 4.

The final answer is \boxed{4}.