Question

Write 3log7x-4log7y as a single logarithm.

Answer 1

Answer:

$3\log7x - 4\log7y = log(\frac{x^3}{7y^4} )$

Step-by-step explanation:

The given logarithm can be re-written as:

$\log(7x)^3 - log(7y)^4$

For the power property:

$\log_ba^c = c\log_ba$

Know for the quotient property you get:

$log_ba - log_bc = log(\frac{a}{c} )$

$\log{\frac{(7x)^3}{(7y)^4} } =\log{\frac{7^3x^3}{7^4y^4} } =\log{\frac{x^3}{7y^4} }$

so the final answer is $\log{\frac{x^3}{7y^4} }$

Answer 2

Answer:

Omaeyou Mu shendeiru, NANI?