Solve for v 6^v<3^v

$6^{v} < 3^{v}$

Take the log of both sides of the inequality.

$\ln (6^{v}) < \ln (3^{v})$

Expand $\ln (6^{v})$ by moving $v$ outside the logarithm.

$v \ln (6) < \ln (3^{v})$

Expand $\ln (3^{v})$ by moving $v$ outside the logarithm.

$v \ln (6) < v \ln (3)$

Solve the inequality for $v$ .

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Subtract $v \ln (3)$ from both sides of the inequality.

$v \ln (6) - v \ln (3) < 0$

Factor $v$ out of $v \ln (6) - v \ln (3)$ .

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Factor $v$ out of $v \ln (6)$ .

$v (\ln (6)) - v \ln (3) < 0$

Factor $v$ out of $- v \ln (3)$ .

$v (\ln (6)) + v (- 1 \ln (3)) < 0$

Factor $v$ out of $v \ln (6) + v (- 1 \ln (3))$ .

$v (\ln (6) - 1 \ln (3)) < 0$

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$v (\ln (6) - \ln (3)) < 0$

Divide each term by $\ln (6) - \ln (3)$ and simplify.

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Convert the inequality to an equation.

$v (\ln (6) - \ln (3)) = 0$

Divide each term in $v (\ln (6) - \ln (3)) = 0$ by $\ln (6) - \ln (3)$ .

$\frac{v (\ln (6) - \ln (3))}{\ln (6) - \ln (3)} = \frac{0}{\ln (6) - \ln (3)}$

Cancel the common factor of $\ln (6) - \ln (3)$ .

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Cancel the common factor.

$\frac{v (\ln (6) - \ln (3))}{\ln (6) - \ln (3)} = \frac{0}{\ln (6) - \ln (3)}$

Divide $v$ by $1$ .

$v = \frac{0}{\ln (6) - \ln (3)}$

Divide $0$ by $\ln (6) - \ln (3)$ .

$v = 0$

The solution consists of all of the true intervals.

$v < 0$

The result can be shown in multiple forms.

Inequality Form:

$v < 0$

Interval Notation:

$(- \infty, 0)$

Solve for v 6^v<3^v

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