Precalculus Examples

$e^{2 (\ln (3))} + 2 e^{\ln (3)} - 15 = 0$

Step 1

Simplify ${(2.71828182)}^{2 (\log_{2.71828182} (3))} + 2 {(2.71828182)}^{\log_{2.71828182} (3)} - 15$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Simplify $2 (\log_{2.71828182} (3))$ by moving $2$ inside the logarithm.

${2.71828182}^{\log_{2.71828182} (3^{2})} + 2 {(2.71828182)}^{\log_{2.71828182} (3)} - 15 = 0$

Step 1.1.2

Exponentiation and log are inverse functions.

$3^{2} + 2 {(2.71828182)}^{\log_{2.71828182} (3)} - 15 = 0$

Step 1.1.3

Raise $3$ to the power of $2$ .

$9 + 2 {(2.71828182)}^{\log_{2.71828182} (3)} - 15 = 0$

Step 1.1.4

Exponentiation and log are inverse functions.

$9 + 2 \cdot 3 - 15 = 0$

Step 1.1.5

Multiply $2$ by $3$ .

$9 + 6 - 15 = 0$

Step 1.2

Simplify by adding and subtracting.

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Step 1.2.1

Add $9$ and $6$ .

$15 - 15 = 0$

Step 1.2.2

Subtract $15$ from $15$ .

$0 = 0$

Step 2

Since $0 = 0$ , the equation will always be true.

Always true

Step 3

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$