Solve for x 38^(2x)=109

$38^{2 x} = 109$

Step 1

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

$\ln (38^{2 x}) = \ln (109)$

Step 2

Expand $\ln (38^{2 x})$ by moving $2 x$ outside the logarithm.

$2 x \ln (38) = \ln (109)$

Step 3

Divide each term in $2 x \ln (38) = \ln (109)$ by $2 \ln (38)$ and simplify.

Divide each term in $2 x \ln (38) = \ln (109)$ by $2 \ln (38)$ .

$\frac{2 x \ln (38)}{2 \ln (38)} = \frac{\ln (109)}{2 \ln (38)}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x \ln (38)}{2 \ln (38)} = \frac{\ln (109)}{2 \ln (38)}$

Rewrite the expression.

$\frac{x \ln (38)}{\ln (38)} = \frac{\ln (109)}{2 \ln (38)}$

Cancel the common factor of $\ln (38)$ .

Cancel the common factor.

$\frac{x \ln (38)}{\ln (38)} = \frac{\ln (109)}{2 \ln (38)}$

Divide $x$ by $1$ .

$x = \frac{\ln (109)}{2 \ln (38)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (109)}{2 \ln (38)}$

Decimal Form:

$x = 0.64484354 \dots$