Show that the solution of the equation $2^{3x+4}\times 5^{2x-1}=16^{x}\times 5^{3x}$ is $\lg\dfrac{16}{5}$. Express $2\log_{2}x-\log_{2}(x-4)=3$ as a quadratic equation in $x$ and explain why there are no real solutions.

1. Express all terms with base 2 and base 5, and group terms with the same base.

$2^{3x+4} \times 5^{2x-1} = (2^4)^x \times 5^{3x}$
$2^{3x+4} \times 5^{2x-1} = 2^{4x} \times 5^{3x}$

2. Rearrange the equation to collect terms with the same base on opposite sides.

$\frac{2^{3x+4}}{2^{4x}} = \frac{5^{3x}}{5^{2x-1}}$
$2^{3x+4-4x} = 5^{3x-(2x-1)}$
$2^{4-x} = 5^{x+1}$

3. Take logarithms on both sides and solve for x.

$\lg(2^{4-x}) = \lg(5^{x+1})$
$(4-x)\lg 2 = (x+1)\lg 5$
$4\lg 2 - x\lg 2 = x\lg 5 + \lg 5$
$4\lg 2 - \lg 5 = x\lg 5 + x\lg 2$
$\lg(2^4) - \lg 5 = x(\lg 5 + \lg 2)$
$\lg 16 - \lg 5 = x\lg(5 \times 2)$
$\lg\left(\frac{16}{5}\right) = x\lg 10$
Since $\lg 10 = 1$,
$x = \lg\frac{16}{5}$

1. Apply the power law of logarithms

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

2. Apply the quotient law of logarithms

$\log_{2}\left(\frac{x^2}{x-4}\right) = 3$

3. Convert the logarithmic equation to exponential form and rearrange into a quadratic equation

$\frac{x^2}{x-4} = 2^3$
$\frac{x^2}{x-4} = 8$
$x^2 = 8(x-4)$
$x^2 - 8x + 32 = 0$

4. Calculate the discriminant of the quadratic equation

$b^2 - 4ac = (-8)^2 - 4(1)(32)$
$= 64 - 128$
$= -64$

5. Conclude based on the value of the discriminant

Since the discriminant $b^2 - 4ac < 0$, the quadratic equation has no real solutions.