What is the answer to solvefor p,pH=-log_{10}(H^+) ?

The answer to solvefor p,pH=-log_{10}(H^+) is p=-(log_{10}(H))/H ;\quad H\ne 0

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solvefor p,pH=-log_{10}(H^+)

solve for

p, p H = - lo g_{10} (H^{+})

p = - \frac{lo g _{10} ( H )}{H}; H \neq = 0

Solution steps

p H = - lo g_{10} (H^{+})

Divide both sides by

H; H \neq = 0

p = - \frac{lo g _{10} ( H )}{H}; H \neq = 0

5 x + 11 = 21

a, - B a + 9 = Q^{2} a

x, 2 x + 9 y = 36

x, 5 x - 8 y = - 29

4 (x - 2) - 5 (2 - 3 x) = 4 (2 x - 6)

What is the answer to solvefor p,pH=-log_{10}(H^+) ?
The answer to solvefor p,pH=-log_{10}(H^+) is p=-(log_{10}(H))/H ;\quad H\ne 0