Solve for x log base 2 of 4^(3x-1)=25^( log base 5 of 2)

$\log_{2} (4^{3 x - 1}) = 25^{\log_{5} (2)}$

Expand $\log_{2} (4^{3 x - 1})$ .

Expand $\log_{2} (4^{3 x - 1})$ by moving $3 x - 1$ outside the logarithm.

$(3 x - 1) \log_{2} (4)$

Logarithm base $2$ of $4$ is $2$ .

$(3 x - 1) \cdot 2$

The expanded equation is $(3 x - 1) \cdot 2 = 25^{\log_{5} (2)}$ .

$(3 x - 1) \cdot 2 = 25^{\log_{5} (2)}$

Divide each term in $((3 x - 1) \cdot 2) = 25^{\log_{5} (2)}$ by $2$ and simplify.

Divide each term in $((3 x - 1) \cdot 2) = 25^{\log_{5} (2)}$ by $2$ .

$\frac{((3 x - 1) \cdot 2)}{2} = \frac{25^{\log_{5} (2)}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{(3 x - 1) \cdot 2}{2} = \frac{25^{\log_{5} (2)}}{2}$

Divide $3 x - 1$ by $1$ .

$3 x - 1 = \frac{25^{\log_{5} (2)}}{2}$

Add $1$ to both sides of the equation.

$3 x = \frac{25^{\log_{5} (2)}}{2} + 1$

Divide each term in $3 x = \frac{25^{\log_{5} (2)}}{2} + 1$ by $3$ and simplify.

Divide each term in $3 x = \frac{25^{\log_{5} (2)}}{2} + 1$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{25^{\log_{5} (2)}}{2}}{3} + \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{25^{\log_{5} (2)}}{2}}{3} + \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{\frac{25^{\log_{5} (2)}}{2}}{3} + \frac{1}{3}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{25^{\log_{5} (2)}}{2} \cdot \frac{1}{3} + \frac{1}{3}$

Combine.

$x = \frac{25^{\log_{5} (2)} \cdot 1}{2 \cdot 3} + \frac{1}{3}$

Multiply $25^{\log_{5} (2)}$ by $1$ .

$x = \frac{25^{\log_{5} (2)}}{2 \cdot 3} + \frac{1}{3}$

Multiply $2$ by $3$ .

$x = \frac{25^{\log_{5} (2)}}{6} + \frac{1}{3}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{25^{\log_{5} (2)}}{6} + \frac{1}{3}$

Decimal Form:

$x = 1$