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If y = [1/3 x3 + 2x + 5]2, find dy/dx. (b) Given that log3 = 0.4771 and log2 = 0.3010, evaluate log162 + log36 - log54.

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2 parts
a

2(13x3+2x+5)(x2+2)2\left(\frac{1}{3}x^3 + 2x + 5\right)(x^2 + 2)

b
2.0333

Step-by-step solution

8 steps

Part a· 3 steps

If y=[13x3+2x+5]2y=\left[\frac{1}{3}x^3 + 2x + 5\right]^2, find dydx\frac{dy}{dx}.

  1. Apply the chain rule.

    Let u=13x3+2x+5u = \frac{1}{3}x^3 + 2x + 5.
    Then y=u2y = u^2.
    The derivative of yy with respect to uu is dydu=2u\frac{dy}{du} = 2u.

  2. Find the derivative of uu with respect to xx.

    dudx=ddx(13x3+2x+5)\frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 + 2x + 5\right) dudx=13(3x3−1)+2(1)+0\frac{du}{dx} = \frac{1}{3}(3x^{3-1}) + 2(1) + 0 dudx=x2+2\frac{du}{dx} = x^2 + 2

  3. Use the chain rule formula dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

    dydx=(2u)(x2+2)\frac{dy}{dx} = (2u)(x^2 + 2)

    Substitute u=13x3+2x+5u = \frac{1}{3}x^3 + 2x + 5 back into the expression.

    dydx=2(13x3+2x+5)(x2+2)\frac{dy}{dx} = 2\left(\frac{1}{3}x^3 + 2x + 5\right)(x^2 + 2)

    The derivative is:

    2(13x3+2x+5)(x2+2)\boxed{2\left(\frac{1}{3}x^3 + 2x + 5\right)(x^2 + 2)}

Part b2.0333

Given that log⁡3=0.4771\log 3 = 0.4771 and log⁡2=0.3010\log 2 = 0.3010, evaluate log⁡162+log⁡36−log⁡54\log 162 + \log 36 - \log 54.

  1. Use the logarithm properties log⁡A+log⁡B=log⁡(AB)\log A + \log B = \log(AB) and log⁡A−log⁡B=log⁡(AB)\log A - \log B = \log\left(\frac{A}{B}\right) to combine the terms.

    log⁡162+log⁡36−log⁡54=log⁡(162×3654)\log 162 + \log 36 - \log 54 = \log\left(\frac{162 \times 36}{54}\right)

  2. Simplify the expression inside the logarithm.

    162×3654=16254×36\frac{162 \times 36}{54} = \frac{162}{54} \times 36

    Since 162=3×54162 = 3 \times 54:

    =3×36= 3 \times 36 =108= 108

    So the expression becomes log⁡108\log 108.

  3. Express 108 as a product of powers of 2 and 3.

    108=2×54=2×2×27=22×33108 = 2 \times 54 = 2 \times 2 \times 27 = 2^2 \times 3^3

  4. Apply logarithm properties log⁡(AB)=log⁡A+log⁡B\log(AB) = \log A + \log B and log⁡(An)=nlog⁡A\log(A^n) = n \log A.

    log⁡108=log⁡(22×33)=log⁡(22)+log⁡(33)\log 108 = \log(2^2 \times 3^3) = \log(2^2) + \log(3^3) =2log⁡2+3log⁡3= 2 \log 2 + 3 \log 3

  5. Substitute the given values for log⁡2\log 2 and log⁡3\log 3.

    =2(0.3010)+3(0.4771)= 2(0.3010) + 3(0.4771) =0.6020+1.4313= 0.6020 + 1.4313 =2.0333= 2.0333

    The evaluated expression is:

    2.0333\boxed{2.0333}
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