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expand solve(0= 1/2 (2)-1/2 ln(2^{1/2})+c,c)

Solution

expand

so l v e (0 = \frac{1}{2} (2) - \frac{1}{2} ln (2^{\frac{1}{2}}) + c, c)

Solution

so l v e (0 = \frac{1}{2} \cdot 2 - \frac{1}{2} \cdot \frac{1}{2} ln (2) + c, c)

Solution steps

so l v e (0 = \frac{1}{2} (2) - \frac{1}{2} ln (2^{\frac{1}{2}}) + c, c)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (2^{\frac{1}{2}}) = \frac{1}{2} ln (2)

= so l v e (0 = \frac{1}{2} \cdot 2 - \frac{1}{2} \cdot \frac{1}{2} ln (2) + c, c)

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Frequently Asked Questions (FAQ)

What is expand solve(0= 1/2 (2)-1/2 ln(2^{1/2})+c,c) ?
The solution to expand solve(0= 1/2 (2)-1/2 ln(2^{1/2})+c,c) is solve(0= 1/2*2-1/2*1/2 ln(2)+c,c)