Solve the equation $15(16^x) - 4^x = 6$. Solve the equation $\lg(2y) + 2\ln(y) = 2$.

1. Express the equation as a quadratic in terms of $4^x$.

Let $u = 4^x$.
Since $16^x = (4^2)^x = (4^x)^2 = u^2$, the equation becomes:
$15u^2 - u = 6$
$15u^2 - u - 6 = 0$

2. Solve the quadratic equation for $u$.

$(5u + 3)(3u - 2) = 0$
$u = -\frac{3}{5} \quad \text{or} \quad u = \frac{2}{3}$

3. Substitute back $u = 4^x$ and reject the invalid solution.

$4^x = -\frac{3}{5} \quad \text{or} \quad 4^x = \frac{2}{3}$
Since $4^x > 0$ for all real $x$, $4^x = -\frac{3}{5}$ has no real solution.

4. Solve for $x$ using logarithms.

$4^x = \frac{2}{3}$
Take $\ln$ on both sides:
$x \ln 4 = \ln \left(\frac{2}{3}\right)$
$x = \frac{\ln(2/3)}{\ln 4}$
$x \approx -0.29248$
$x = -0.292 \text{ (to 3 s.f.)}$

1. Apply logarithm laws to separate the terms in $\lg(2y)$

$\lg 2 + \lg y + 2\ln y = 2$

2. Change the base of $\lg y$ to base $e$ (or vice versa) to combine the terms involving $y$

$\lg 2 + \frac{\ln y}{\ln 10} + 2\ln y = 2$
$\ln y \left( \frac{1}{\ln 10} + 2 \right) = 2 - \lg 2$

3. Solve for $y$

$\ln y = \frac{2 - \lg 2}{\frac{1}{\ln 10} + 2}$
$y = e^{\frac{2 - \lg 2}{\frac{1}{\ln 10} + 2}}$
$y \approx 1.91$ (to 3 s.f.)