Precalculus Examples

$- 7 i \cdot (3 i) = 21 i^{2}$

Step 1

Simplify.

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Step 1.1

Simplify both sides of the equation.

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Step 1.1.1

Multiply $3$ by $- 7$ .

$- 21 i \cdot i = 21 i^{2}$

Step 1.1.2

Raise $i$ to the power of $1$ .

$- 21 (i^{1} i) = 21 i^{2}$

Step 1.1.3

Raise $i$ to the power of $1$ .

$- 21 (i^{1} i^{1}) = 21 i^{2}$

Step 1.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 21 i^{1 + 1} = 21 i^{2}$

Step 1.1.5

Add $1$ and $1$ .

$- 21 i^{2} = 21 i^{2}$

Step 1.2

Divide each term in $- 21 i^{2} = 21 i^{2}$ by $21$ .

$\frac{- 21 i^{2}}{21} = \frac{21 i^{2}}{21}$

Step 1.3

Rewrite $i^{2}$ as $- 1$ .

$\frac{21 \cdot - 1}{21}$

Step 1.4

Multiply $21$ by $- 1$ .

$\frac{- 21}{21}$

Step 1.5

Divide $- 21$ by $21$ .

$- 1$

Step 1.6

Reduce by cancelling the common factors.

$i^{2} = - 1$

Step 2

Convert the exponential equation to a logarithmic equation using the logarithm base $(i)$ of the right side $(- 1)$ equals the exponent $(2)$ .

$\log_{i} (- 1) = 2$