The value of $\left(\frac{1}{3}\log_{10} 125 - 2\log_{10} 4 + \log_{10} 32\right)$ is

### Concept & Formula This problem requires consolidating multiple logarithmic terms into a single logarithm using the power, product, and quotient rules. The Power Rule: $$ \log_a(x^n) = n \cdot \log_a(x) $$ The Product Rule: $$ \log_a(xy) = \log_a(x) + \log_a(y) $$ The Quotient Rule: $$ \log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y) $$ ### Step-by-Step Solution Let's simplify each term individually using the power rule to bring the coefficient into the exponent. First term: $\frac{1}{3}\log_{10}(125)$ Since $125 = 5^3$, we have: $\frac{1}{3}\log_{10}(5^3) = \log_{10}((5^3)^{1/3}) = \log_{10}(5^1) = \log_{10}(5)$ Second term: $2\log_{10}(4)$ Bring the $2$ up as a power: $\log_{10}(4^2) = \log_{10}(16)$ Third term remains as is: $\log_{10}(32)$ Now substitute these simplified terms back into the original expression: $\log_{10}(5) - \log_{10}(16) + \log_{10}(32)$ Apply the product and quotient rules to combine them into a single log: $\log_{10}\left(\frac{5 \times 32}{16}\right)$ Simplify the fraction: $\frac{5 \times 32}{16} = 5 \times 2 = 10$ So the expression becomes: $\log_{10}(10)$ Since $\log_a(a) = 1$, we get: $\log_{10}(10) = 1$ ### Exam Strategy & Shortcut Instead of combining into a single large number, convert everything to prime bases immediately. $125 = 5^3$, $4 = 2^2$, $32 = 2^5$. Expression = $(1/3) \cdot 3\log_{10}(5) - 2 \cdot 2\log_{10}(2) + 5\log_{10}(2)$ $= \log_{10}(5) - 4\log_{10}(2) + 5\log_{10}(2)$ $= \log_{10}(5) + \log_{10}(2) = \log_{10}(10) = 1$. This avoids fraction multiplication entirely. ### Common Pitfall A frequent mistake is mishandling the power rule with fractional coefficients. Students might incorrectly evaluate $(1/3)\log(125)$ by dividing the entire log value rather than applying the cube root to the argument $125$. Always apply the exponent to the argument first. ### Final Answer **Therefore, the correct answer is 1.**