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2 years ago

How do you solve log base 2 of 0.5 power 3 ?

Mathematics

1 answer:

Vinil7 [7]2 years ago

7 0

Answer: - 3

Step-by-step explanation: please see attachment for explanation

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Find dy/dx if y =x^3+5x+2/x²-1

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<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

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How do you solve log base 2 of 0.5 power 3 ?​

How do you solve log base 2 of 0.5 power 3 ?