Solve for x 3^(2x-1)=7

$3^{2 x - 1} = 7$

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

$\ln (3^{2 x - 1}) = \ln (7)$

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

$(2 x - 1) \ln (3) = \ln (7)$

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

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Apply the distributive property.

$2 x \ln (3) - 1 \ln (3) = \ln (7)$

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

$2 x \ln (3) - \ln (3) = \ln (7)$

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

$2 x \ln (3) - \ln (3) - \ln (7) = 0$

Move all terms not containing $x$ to the right side of the equation.

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Add $\ln (3)$ to both sides of the equation.

$2 x \ln (3) - \ln (7) = \ln (3)$

Add $\ln (7)$ to both sides of the equation.

$2 x \ln (3) = \ln (3) + \ln (7)$

Divide each term by $2 \ln (3)$ and simplify.

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Divide each term in $2 x \ln (3) = \ln (3) + \ln (7)$ by $2 \ln (3)$ .

$\frac{2 x \ln (3)}{2 \ln (3)} = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

Simplify $\frac{2 x \ln (3)}{2 \ln (3)}$ .

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x \ln (3)}{2 \ln (3)} = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

Rewrite the expression.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

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

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Cancel the common factor.

$\frac{x \ln (3)}{\ln (3)} = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

Divide $x$ by $1$ .

$x = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

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

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Cancel the common factor.

$x = \frac{\ln (3)}{2 \ln (3)} + \frac{\ln (7)}{2 \ln (3)}$

Rewrite the expression.

$x = \frac{1}{2} + \frac{\ln (7)}{2 \ln (3)}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2} + \frac{\ln (7)}{2 \ln (3)}$

Decimal Form:

$x = 1.38562187 \dots$

Solve for x 3^(2x-1)=7

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