Finite Math Examples

$(2 - m) = \log (5 c + 5)$

Step 1

Rewrite the equation as $\log (5 c + 5) = 2 - m$ .

$\log (5 c + 5) = 2 - m$

Step 2

Rewrite $\log (5 c + 5) = 2 - m$ 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 $b^{y} = x$ .

$10^{2 - m} = 5 c + 5$

Step 3

Solve for $c$ .

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Step 3.1

Rewrite the equation as $5 c + 5 = 10^{2 - m}$ .

$5 c + 5 = 10^{2 - m}$

Step 3.2

Subtract $5$ from both sides of the equation.

$5 c = 10^{2 - m} - 5$

Step 3.3

Divide each term in $5 c = 10^{2 - m} - 5$ by $5$ and simplify.

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Step 3.3.1

Divide each term in $5 c = 10^{2 - m} - 5$ by $5$ .

$\frac{5 c}{5} = \frac{10^{2 - m}}{5} + \frac{- 5}{5}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $5$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{5 c}{5} = \frac{10^{2 - m}}{5} + \frac{- 5}{5}$

Step 3.3.2.1.2

Divide $c$ by $1$ .

$c = \frac{10^{2 - m}}{5} + \frac{- 5}{5}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $- 5$ by $5$ .

$c = \frac{10^{2 - m}}{5} - 1$

Step 3.3.3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$c = \frac{10^{2 - m}}{5} - 1 \cdot \frac{5}{5}$

Step 3.3.3.3

Combine $- 1$ and $\frac{5}{5}$ .

$c = \frac{10^{2 - m}}{5} + \frac{- 1 \cdot 5}{5}$

Step 3.3.3.4

Combine the numerators over the common denominator.

$c = \frac{10^{2 - m} - 1 \cdot 5}{5}$

Step 3.3.3.5

Multiply $- 1$ by $5$ .

$c = \frac{10^{2 - m} - 5}{5}$