Solve for x 4^(2x)*2^x=476

$4^{2 x} \cdot 2^{x} = 476$

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

$\ln (4^{2 x} \cdot 2^{x}) = \ln (476)$

Expand the left side.

Rewrite $\ln (4^{2 x} \cdot 2^{x})$ as $\ln (4^{2 x}) + \ln (2^{x})$ .

$\ln (4^{2 x}) + \ln (2^{x}) = \ln (476)$

Expand $\ln (4^{2 x})$ by moving $2 x$ outside the logarithm.

$2 x \ln (4) + \ln (2^{x}) = \ln (476)$

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

$2 x \ln (4) + x \ln (2) = \ln (476)$

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

$2 x \ln (4) + x \ln (2) - \ln (476) = 0$

Add $\ln (476)$ to both sides of the equation.

$2 x \ln (4) + x \ln (2) = \ln (476)$

Factor $x$ out of $2 x \ln (4) + x \ln (2)$ .

Factor $x$ out of $2 x \ln (4)$ .

$x (2 \ln (4)) + x \ln (2) = \ln (476)$

Factor $x$ out of $x \ln (2)$ .

$x (2 \ln (4)) + x (\ln (2)) = \ln (476)$

Factor $x$ out of $x (2 \ln (4)) + x (\ln (2))$ .

$x (2 \ln (4) + \ln (2)) = \ln (476)$

Divide each term in $x (2 \ln (4) + \ln (2)) = \ln (476)$ by $2 \ln (4) + \ln (2)$ and simplify.

Divide each term in $x (2 \ln (4) + \ln (2)) = \ln (476)$ by $2 \ln (4) + \ln (2)$ .

$\frac{x (2 \ln (4) + \ln (2))}{2 \ln (4) + \ln (2)} = \frac{\ln (476)}{2 \ln (4) + \ln (2)}$

Simplify the left side.

Cancel the common factor of $2 \ln (4) + \ln (2)$ .

Cancel the common factor.

$\frac{x (2 \ln (4) + \ln (2))}{2 \ln (4) + \ln (2)} = \frac{\ln (476)}{2 \ln (4) + \ln (2)}$

Divide $x$ by $1$ .

$x = \frac{\ln (476)}{2 \ln (4) + \ln (2)}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (476)}{2 \ln (4) + \ln (2)}$

Decimal Form:

$x = 1.77896355 \dots$