Calculus Examples

$\ln (\frac{x^{2} (3 x - 5)}{3 (2 x + 1) e^{7 x}})$

Step 1

Rewrite $\ln (\frac{x^{2} (3 x - 5)}{3 (2 x + 1) e^{7 x}})$ as $\ln (x^{2} (3 x - 5)) - \ln (3 (2 x + 1) e^{7 x})$ .

$\ln (x^{2} (3 x - 5)) - \ln (3 (2 x + 1) e^{7 x})$

Step 2

Rewrite $\ln (x^{2} (3 x - 5))$ as $\ln (x^{2}) + \ln (3 x - 5)$ .

$\ln (x^{2}) + \ln (3 x - 5) - \ln (3 (2 x + 1) e^{7 x})$

Step 3

Expand $\ln (x^{2})$ by moving $2$ outside the logarithm.

$2 \ln (x) + \ln (3 x - 5) - \ln (3 (2 x + 1) e^{7 x})$

Step 4

Rewrite $\ln (3 (2 x + 1) e^{7 x})$ as $\ln (3 (2 x + 1)) + \ln (e^{7 x})$ .

$2 \ln (x) + \ln (3 x - 5) - (\ln (3 (2 x + 1)) + \ln (e^{7 x}))$

Step 5

Rewrite $\ln (3 (2 x + 1))$ as $\ln (3) + \ln (2 x + 1)$ .

$2 \ln (x) + \ln (3 x - 5) - (\ln (3) + \ln (2 x + 1) + \ln (e^{7 x}))$

Step 6

Expand $\ln (e^{7 x})$ by moving $7 x$ outside the logarithm.

$2 \ln (x) + \ln (3 x - 5) - (\ln (3) + \ln (2 x + 1) + 7 x \ln (e))$

Step 7

The natural logarithm of $e$ is $1$ .

$2 \ln (x) + \ln (3 x - 5) - (\ln (3) + \ln (2 x + 1) + 7 x \cdot 1)$

Step 8

Multiply $7$ by $1$ .

$2 \ln (x) + \ln (3 x - 5) - (\ln (3) + \ln (2 x + 1) + 7 x)$

Step 9

Simplify each term.

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Step 9.1

Apply the distributive property.

$2 \ln (x) + \ln (3 x - 5) - \ln (3) - \ln (2 x + 1) - (7 x)$

Step 9.2

Multiply $7$ by $- 1$ .

$2 \ln (x) + \ln (3 x - 5) - \ln (3) - \ln (2 x + 1) - 7 x$