Solve for O 7.15=- log of H_3O

$7.15 = - \log (H_{3} O)$

Step 1

Rewrite the equation as $- \log (H_{3} O) = 7.15$ .

$- \log (H_{3} O) = 7.15$

Step 2

Divide each term in $- \log (H_{3} O) = 7.15$ by $- 1$ and simplify.

Divide each term in $- \log (H_{3} O) = 7.15$ by $- 1$ .

$\frac{- \log (H_{3} O)}{- 1} = \frac{7.15}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\log (H_{3} O)}{1} = \frac{7.15}{- 1}$

Divide $\log (H_{3} O)$ by $1$ .

$\log (H_{3} O) = \frac{7.15}{- 1}$

Simplify the right side.

Divide $7.15$ by $- 1$ .

$\log (H_{3} O) = - 7.15$

Step 3

Rewrite $\log (H_{3} O) = - 7.15$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$10^{- 7.15} = H_{3} O$

Step 4

Solve for $O$ .

Rewrite the equation as $H_{3} O = 10^{- 7.15}$ .

$H_{3} O = 10^{- 7.15}$

Divide each term in $(H_{3} O) = 10^{- 7.15}$ by $H_{3}$ and simplify.

Divide each term in $(H_{3} O) = 10^{- 7.15}$ by $H_{3}$ .

$\frac{(H_{3} O)}{H_{3}} = \frac{10^{- 7.15}}{H_{3}}$

Simplify the left side.

Cancel the common factor of $H_{3}$ .

Cancel the common factor.

$\frac{H_{3} O}{H_{3}} = \frac{10^{- 7.15}}{H_{3}}$

Divide $O$ by $1$ .

$O = \frac{10^{- 7.15}}{H_{3}}$

Simplify the right side.

Simplify the expression.

Move $10^{- 7.15}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$O = \frac{1}{H_{3} \cdot 10^{7.15}}$

Raise $10$ to the power of $7.15$ .

$O = \frac{1}{H_{3} \cdot 14125375.4462276}$

Move $14125375.4462276$ to the left of $H_{3}$ .

$O = \frac{1}{14125375.4462276 \cdot H_{3}}$

Rewrite $1$ as $1 (1)$ .

$O = \frac{1 (1)}{14125375.4462276 \cdot H_{3}}$

Factor $14125375.4462276$ out of $14125375.4462276 \cdot H_{3}$ .

$O = \frac{1 (1)}{14125375.4462276 (H_{3})}$

Separate fractions.

$O = \frac{1}{14125375.4462276} \cdot \frac{1}{H_{3}}$

Divide $1$ by $14125375.4462276$ .

$O = 0.00000007 \frac{1}{H_{3}}$

Combine $0.00000007$ and $\frac{1}{H_{3}}$ .

$O = \frac{0.00000007}{H_{3}}$

Solve for O 7.15=- log of H_3O

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