Given that lg 2 = m, express in terms of m, $\log_{5}32$ $2^{10}$ Express $2\log_{3}x - \log_{3}2 = \log_{3}(x - 4)$ as a quadratic equation in $x$ and explain why the quadratic equation has no real solutions.

1. Change the base of the logarithm to 10

$\log_{5}32 = \frac{\lg 32}{\lg 5}$

2. Express the arguments in terms of 2 and 10

$\frac{\lg 32}{\lg 5} = \frac{\lg (2^5)}{\lg (\frac{10}{2})}$

3. Apply logarithm laws and substitute the given value

$\frac{5 \lg 2}{\lg 10 - \lg 2} = \frac{5m}{1 - m}$

1. Express 2 in exponential form using the given logarithmic equation

Since $\lg 2 = m$, we have $\log_{10} 2 = m$, which means $2 = 10^m$.

2. Substitute and simplify to find $2^{10}$

$2^{10} = (10^m)^{10} = 10^{10m}$

1. Apply the power law of logarithms

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

2. Apply the quotient law of logarithms

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

3. Equate the arguments and rearrange into a quadratic equation

$\frac{x^2}{2} = x - 4$
$x^2 = 2x - 8$
$x^2 - 2x + 8 = 0$

4. Calculate the discriminant to show there are no real solutions

For $x^2 - 2x + 8 = 0$, the discriminant is:
$b^2 - 4ac = (-2)^2 - 4(1)(8)$
$= 4 - 32 = -28$
Since the discriminant is less than $0$, the quadratic equation has no real solutions.