Solve by Factoring 2/3*(4^(3x))-5=-2

$\frac{2}{3} \cdot (4^{3 x}) - 5 = - 2$

Step 1

Add $2$ to both sides of the equation.

$\frac{2}{3} \cdot (4^{3 x}) - 5 + 2 = 0$

Step 2

Simplify $\frac{2}{3} \cdot (4^{3 x}) - 5 + 2$ .

Multiply $\frac{2}{3} (4^{3 x})$ .

Combine $\frac{2}{3}$ and $4^{3 x}$ .

$\frac{2 \cdot 4^{3 x}}{3} - 5 + 2 = 0$

Rewrite $4$ as $2^{2}$ .

$\frac{2 \cdot {(2^{2})}^{3 x}}{3} - 5 + 2 = 0$

Multiply the exponents in ${(2^{2})}^{3 x}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{2 \cdot 2^{2 (3 x)}}{3} - 5 + 2 = 0$

Multiply $3$ by $2$ .

$\frac{2 \cdot 2^{6 x}}{3} - 5 + 2 = 0$

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

$\frac{2^{1 + 6 x}}{3} - 5 + 2 = 0$

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2^{1 + 6 x}}{3} - 5 \cdot \frac{3}{3} + 2 = 0$

Combine $- 5$ and $\frac{3}{3}$ .

$\frac{2^{1 + 6 x}}{3} + \frac{- 5 \cdot 3}{3} + 2 = 0$

Combine the numerators over the common denominator.

$\frac{2^{1 + 6 x} - 5 \cdot 3}{3} + 2 = 0$

Multiply $- 5$ by $3$ .

$\frac{2^{1 + 6 x} - 15}{3} + 2 = 0$

Step 3

To write $2$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2^{1 + 6 x} - 15}{3} + 2 \cdot \frac{3}{3} = 0$

Step 4

Combine $2$ and $\frac{3}{3}$ .

$\frac{2^{1 + 6 x} - 15}{3} + \frac{2 \cdot 3}{3} = 0$

Step 5

Combine the numerators over the common denominator.

$\frac{2^{1 + 6 x} - 15 + 2 \cdot 3}{3} = 0$

Step 6

Simplify the numerator.

Multiply $2$ by $3$ .

$\frac{2^{1 + 6 x} - 15 + 6}{3} = 0$

Add $- 15$ and $6$ .

$\frac{2^{1 + 6 x} - 9}{3} = 0$

Step 7

Set the numerator equal to zero.

$2^{1 + 6 x} - 9 = 0$

Step 8

Solve the equation for $x$ .

Add $9$ to both sides of the equation.

$2^{1 + 6 x} = 9$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{1 + 6 x}) = \ln (9)$

Expand $\ln (2^{1 + 6 x})$ by moving $1 + 6 x$ outside the logarithm.

$(1 + 6 x) \ln (2) = \ln (9)$

Simplify the left side.

Simplify $(1 + 6 x) \ln (2)$ .

Apply the distributive property.

$1 \ln (2) + 6 x \ln (2) = \ln (9)$

Multiply $\ln (2)$ by $1$ .

$\ln (2) + 6 x \ln (2) = \ln (9)$

Reorder $\ln (2)$ and $6 x \ln (2)$ .

$6 x \ln (2) + \ln (2) = \ln (9)$

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

$6 x \ln (2) + \ln (2) - \ln (9) = 0$

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

$6 x \ln (2) + \ln (\frac{2}{9}) = 0$

Subtract $\ln (\frac{2}{9})$ from both sides of the equation.

$6 x \ln (2) = - \ln (\frac{2}{9})$

Divide each term in $6 x \ln (2) = - \ln (\frac{2}{9})$ by $6 \ln (2)$ and simplify.

Divide each term in $6 x \ln (2) = - \ln (\frac{2}{9})$ by $6 \ln (2)$ .

$\frac{6 x \ln (2)}{6 \ln (2)} = \frac{- \ln (\frac{2}{9})}{6 \ln (2)}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x \ln (2)}{6 \ln (2)} = \frac{- \ln (\frac{2}{9})}{6 \ln (2)}$

Rewrite the expression.

$\frac{x \ln (2)}{\ln (2)} = \frac{- \ln (\frac{2}{9})}{6 \ln (2)}$

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{- \ln (\frac{2}{9})}{6 \ln (2)}$

Divide $x$ by $1$ .

$x = \frac{- \ln (\frac{2}{9})}{6 \ln (2)}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{\ln (\frac{2}{9})}{6 \ln (2)}$

Step 9

Simplify $- \frac{\ln (\frac{2}{9})}{6 \ln (2)}$ .

Simplify $6 \ln (2)$ by moving $6$ inside the logarithm.

$x = - \frac{\ln (\frac{2}{9})}{\ln (2^{6})}$

Raise $2$ to the power of $6$ .

$x = - \frac{\ln (\frac{2}{9})}{\ln (64)}$

Solve by Factoring 2/3*(4^(3x))-5=-2

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