Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Answer:
C.
[tex]log( {5^6} \sqrt[3]{12} ) [/tex]
Step-by-step explanation:
6log5 + 1/3 log12
Applying Power law of logarithms
[tex] = log( {5}^{6} ) + log( {12}^{ \frac{1}{3} } ) [/tex]
By applying Fraction index law of exponential
[tex] = log( {5^6} \sqrt[3]{12} ) [/tex]
Answer:
[tex]\textsf{C)}\quad \log \left(5^6 \sqrt[3]{12}\right)[/tex]
Step-by-step explanation:
Given logarithmic expression:
[tex]6 \log 5 + \dfrac{\log 12}{3}[/tex]
Dividing the logarithm of a number by a constant is equivalent to multiplying the logarithm of that number by the reciprocal of the constant. Therefore, the original expression can be rewritten as:
[tex]6 \log 5 + \dfrac{1}{3}\log 12[/tex]
Apply the power rule of logarithms, which states that the logarithm of a number raised to an exponent is equal to the exponent multiplied by the logarithm of the base number:
[tex]\log 5^6 + \log 12^\frac{1}{3}[/tex]
Apply the the fractional exponent rule to the argument of the second term:
[tex]\log 5^6 + \log \sqrt[3]{12}[/tex]
Finally, apply the product rule of logarithms, which states that the logarithm of a product is equal to the sum of the logarithms of the factors:
[tex]\log \left(5^6 \sqrt[3]{12}\right)[/tex]
Therefore, the given expression condensed into a single logarithm is:
[tex]\Large\boxed{\boxed{\log \left(5^6 \sqrt[3]{12}\right)}}[/tex]
[tex]\dotfill[/tex]
Rules used
[tex]\boxed{\begin{array}{c}\underline{\textsf{Power Rule of Logarithms}}\\\\\large\text{$\log x^n=n\log x$}\end{array}}[/tex]
[tex]\boxed{\begin{array}{c}\underline{\textsf{Fractional Exponent Rule}}\\\\\Large\text{$a^{\frac{m}{n}}=\sqrt[n]{a^m}$}\end{array}}[/tex]
[tex]\boxed{\begin{array}{c}\underline{\textsf{Product Rule of Logarithms}}\\\\\large\text{$\log xy=\log x + \log y$}\end{array}}[/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.