Difference between revisions of "Talk:Calculus"

Revision as of 12:36, 5 October 2021

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

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

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

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

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

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

$16(x^{2}+4)=0$

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 <math> log_4 (x^2 + 4x) - log_4 (x^4 + 8x^3 + 16x^2) = 2 </math>
-#Thirde law <math>log_A - log_B = log_({A \over B})</math>
 <math>(x^2 + 4x) = 16(x^4 + 8x^3 + 16x^2)  </math>
 <math> 16 (x^2 + 4)=0 </math>