Express the following as a single logarithm :

$2 \log${10} \space \Big(\dfrac{11}{13}\Big) + \log{10} \space \Big(\dfrac{130}{77}\Big) − \log_{10} \space \Big(\dfrac{55}{91}\Big)

Given,

$\Rightarrow 2 \log${10} \space \Big(\dfrac{11}{13}\Big) + \log{10} \space \Big(\dfrac{130}{77}\Big) − \log_{10} \space \Big(\dfrac{55}{91}\Big)

⇒ 2(log₁₀ 11 - log₁₀ 13) + (log₁₀ 130 - log₁₀77) - (log₁₀ 55 - log₁₀ 91)

⇒ 2log₁₀ 11 - 2log₁₀ 13 + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 11² - log₁₀ 13² + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 121 - log₁₀ 169 + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 121 + log₁₀ 130 + log₁₀ 91 - log₁₀ 169 - log₁₀77 - log₁₀ 55

⇒ log₁₀ (121 × 130 × 91) - log₁₀ (169 × 77 × 55)

⇒ $\log_{10} \space {\dfrac{121 \times 130 \times 91}{169 \times 77 \times 55}}$

⇒ $\log_{10} \space {\dfrac{26}{13}}$

⇒ log₁₀ 2.

Hence, $2 \log${10} \space \Big(\dfrac{11}{13}\Big) + \log{10} \space \Big(\dfrac{130}{77}\Big) − \log_{10} \space \Big(\dfrac{55}{91}\Big) = log₁₀ 2.