What is the answer to solvefor g,g^{-1}(k+1)=2k ?

The answer to solvefor g,g^{-1}(k+1)=2k is g=(k+1)/(2k);\quad k\ne 0

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solvefor g,g^{-1}(k+1)=2k

solve for

g, g^{- 1} (k + 1) = 2 k

g = \frac{k + 1}{2 k}; k \neq = 0

Solution steps

g^{- 1} (k + 1) = 2 k

Refine

\frac{k + 1}{g} = 2 k

Multiply both sides by

g

k + 1 = 2 k g

Switch sides

2 k g = k + 1

Divide both sides by

2 k; k \neq = 0

g = \frac{k + 1}{2 k}; k \neq = 0

2 x - \frac{16000}{x ^{2}} = 0

\frac{x}{8} = \frac{2}{x}

R, n = \frac{P V}{RT}

\frac{z - 6}{2 z + 1} = 10

\frac{1}{x} = \frac{x}{1 - x}

What is the answer to solvefor g,g^{-1}(k+1)=2k ?
The answer to solvefor g,g^{-1}(k+1)=2k is g=(k+1)/(2k);\quad k\ne 0