Talk:Calculus

Need to work on

example 6

ex6: $log_{4}x+log_{4}(x+4)-log_{4}(x^{4}+8x^{3}+16x^{2})=2$ simplify

$log_{4}x+log_{4}(x+4)-log_{4}(x^{4}+8x^{3}+16x^{2})=2$

$log_{4}x+log_{4}(x+4)=2+log_{4}(x^{4}+8x^{3}+16x^{2})$

1. Second law $log_{n}A+log_{n}B=log_{(}AB)$

$log_{4}x+log_{4}(x+4)=log_{4}(x^{2}+4x)$

$log_{4}(x^{2}+4x)-log_{4}(x^{4}+8x^{3}+16x^{2})=2$

1. Thirde law $log_{A}-log_{B}=log_{(}{A \over B})$

$log_{4}(x^{2}+4x)-log_{4}(x^{4}+8x^{3}+16x^{2})=log_{4}({x^{2}+4x \over x^{4}+8x^{3}+16x^{2}})=2$

$log_{4}({1 \over x^{2}+4x})=2$