Solve for x 12^(3x+4)<=2^(5x+3)*3^(2x-1)

$12^{3 x + 4} \leq 2^{5 x + 3} \cdot 3^{2 x - 1}$

Step 1

Take the log of both sides of the inequality.

$\ln (12^{3 x + 4}) \leq \ln (2^{5 x + 3} \cdot 3^{2 x - 1})$

Step 2

Expand $\ln (12^{3 x + 4})$ by moving $3 x + 4$ outside the logarithm.

$(3 x + 4) \ln (12) \leq \ln (2^{5 x + 3} \cdot 3^{2 x - 1})$

Step 3

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

$(3 x + 4) \ln (12) \leq \ln (2^{5 x + 3}) + \ln (3^{2 x - 1})$

Step 4

Expand $\ln (2^{5 x + 3})$ by moving $5 x + 3$ outside the logarithm.

$(3 x + 4) \ln (12) \leq (5 x + 3) \ln (2) + \ln (3^{2 x - 1})$

Step 5

Expand $\ln (3^{2 x - 1})$ by moving $2 x - 1$ outside the logarithm.

$(3 x + 4) \ln (12) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Step 6

Solve the inequality for $x$ .

Simplify the left side.

Simplify $(3 x + 4) \ln (12)$ .

Apply the distributive property.

$3 x \ln (12) + 4 \ln (12) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Simplify $3 \ln (12)$ by moving $3$ inside the logarithm.

$x \ln (12^{3}) + 4 \ln (12) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Simplify $4 \ln (12)$ by moving $4$ inside the logarithm.

$x \ln (12^{3}) + \ln (12^{4}) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Simplify each term.

Raise $12$ to the power of $3$ .

$x \ln (1728) + \ln (12^{4}) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Raise $12$ to the power of $4$ .

$x \ln (1728) + \ln (20736) \leq (5 x + 3) \ln (2) + (2 x - 1) \ln (3)$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$x \ln (1728) + \ln (20736) \leq 5 x \ln (2) + 3 \ln (2) + (2 x - 1) \ln (3)$

Simplify $5 \ln (2)$ by moving $5$ inside the logarithm.

$x \ln (1728) + \ln (20736) \leq x \ln (2^{5}) + 3 \ln (2) + (2 x - 1) \ln (3)$

Simplify $3 \ln (2)$ by moving $3$ inside the logarithm.

$x \ln (1728) + \ln (20736) \leq x \ln (2^{5}) + \ln (2^{3}) + (2 x - 1) \ln (3)$

Simplify each term.

Raise $2$ to the power of $5$ .

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (2^{3}) + (2 x - 1) \ln (3)$

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

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (8) + (2 x - 1) \ln (3)$

Apply the distributive property.

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (8) + 2 x \ln (3) - 1 \ln (3)$

Simplify $2 \ln (3)$ by moving $2$ inside the logarithm.

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (8) + x \ln (3^{2}) - 1 \ln (3)$

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

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (8) + x \ln (3^{2}) - \ln (3)$

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

$x \ln (1728) + \ln (20736) \leq x \ln (32) + \ln (8) + x \ln (9) - \ln (3)$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (1728) + \ln (20736) \leq x \ln (32) + x \ln (9) + \ln (\frac{8}{3})$

Move all the terms containing a logarithm to the left side of the equation.

$x \ln (1728) + \ln (20736) - x \ln (32) - x \ln (9) - \ln (\frac{8}{3}) = 0$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (\frac{20736}{\frac{8}{3}}) = 0$

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (20736 (\frac{3}{8})) = 0$

Cancel the common factor of $8$ .

Factor $8$ out of $20736$ .

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (8 (2592) (\frac{3}{8})) = 0$

Cancel the common factor.

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (8 \cdot (2592 (\frac{3}{8}))) = 0$

Rewrite the expression.

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (2592 \cdot 3) = 0$

Multiply $2592$ by $3$ .

$x \ln (1728) - x \ln (32) - x \ln (9) + \ln (7776) = 0$

Subtract $\ln (7776)$ from both sides of the equation.

$x \ln (1728) - x \ln (32) - x \ln (9) = - \ln (7776)$

Factor $x$ out of $x \ln (1728) - x \ln (32) - x \ln (9)$ .

Factor $x$ out of $x \ln (1728)$ .

$x (\ln (1728)) - x \ln (32) - x \ln (9) = - \ln (7776)$

Factor $x$ out of $- x \ln (32)$ .

$x (\ln (1728)) + x (- 1 \ln (32)) - x \ln (9) = - \ln (7776)$

Factor $x$ out of $- x \ln (9)$ .

$x (\ln (1728)) + x (- 1 \ln (32)) + x (- 1 \ln (9)) = - \ln (7776)$

Factor $x$ out of $x (\ln (1728)) + x (- 1 \ln (32))$ .

$x (\ln (1728) - 1 \ln (32)) + x (- 1 \ln (9)) = - \ln (7776)$

Factor $x$ out of $x (\ln (1728) - 1 \ln (32)) + x (- 1 \ln (9))$ .

$x (\ln (1728) - 1 \ln (32) - 1 \ln (9)) = - \ln (7776)$

Rewrite $- 1 \ln (32)$ as $- \ln (32)$ .

$x (\ln (1728) - \ln (32) - 1 \ln (9)) = - \ln (7776)$

Rewrite $- 1 \ln (9)$ as $- \ln (9)$ .

$x (\ln (1728) - \ln (32) - \ln (9)) = - \ln (7776)$

Divide each term in $x (\ln (1728) - \ln (32) - \ln (9)) = - \ln (7776)$ by $\ln (1728) - \ln (32) - \ln (9)$ and simplify.

Divide each term in $x (\ln (1728) - \ln (32) - \ln (9)) = - \ln (7776)$ by $\ln (1728) - \ln (32) - \ln (9)$ .

$\frac{x (\ln (1728) - \ln (32) - \ln (9))}{\ln (1728) - \ln (32) - \ln (9)} = \frac{- \ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Simplify the left side.

Cancel the common factor of $\ln (1728) - \ln (32) - \ln (9)$ .

Cancel the common factor.

$\frac{x (\ln (1728) - \ln (32) - \ln (9))}{\ln (1728) - \ln (32) - \ln (9)} = \frac{- \ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Step 7

The solution consists of all of the true intervals.

$x \leq - \frac{\ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{\ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}$

Interval Notation:

$(- \infty, - \frac{\ln (7776)}{\ln (1728) - \ln (32) - \ln (9)}]$

Step 9

Solve for x 12^(3x+4)<=2^(5x+3)*3^(2x-1)

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