Answer: (a) \(x=\frac45\). (b) \(y=\frac43\).

(a) First write both logarithms in base \(5\):

\(\log_{25}x=\dfrac{\log_5x}{\log_525}=\dfrac12\log_5x.\)

The equation becomes

\(\log_5(5x-2)-\dfrac12\log_5x=\dfrac12.\)

Multiplying by \(2\),

\(2\log_5(5x-2)-\log_5x=1.\)

Use the laws of logarithms:

\(\log_5\left(\dfrac{(5x-2)^2}{x}\right)=1.\)

Therefore

\(\dfrac{(5x-2)^2}{x}=5.\)

Multiplying by \(x\),

\((5x-2)^2=5x.\)

Expanding,

\(25x^2-20x+4=5x,\)

so

\(25x^2-25x+4=0.\)

Using the quadratic formula,

\(x=\dfrac{25\pm\sqrt{625-400}}{50}=\dfrac{25\pm15}{50}.\)

So

\(x=\dfrac45\quad\text{or}\quad x=\dfrac15.\)

However, \(\log_5(5x-2)\) requires \(5x-2\gt0\), so \(x\gt\dfrac25\). Therefore \(x=\dfrac15\) is rejected, leaving

\(x=\dfrac45.\)

(b) The equation is

\(\mathrm e^{3y-7}+\dfrac4{\mathrm e^3}=\dfrac5{\mathrm e^{3y-1}}.\)

Let

\(z=\mathrm e^{3y-4}.\)

Then

\(\mathrm e^{3y-7}=\dfrac{\mathrm e^{3y-4}}{\mathrm e^3}=\dfrac z{\mathrm e^3},\)

and

\(\mathrm e^{3y-1}=\mathrm e^{3y-4}\mathrm e^3=z\mathrm e^3.\)

So the equation becomes

\(\dfrac z{\mathrm e^3}+\dfrac4{\mathrm e^3}=\dfrac5{z\mathrm e^3}.\)

Multiplying by \(\mathrm e^3\),

\(z+4=\dfrac5z.\)

Multiplying by \(z\),

\(z^2+4z=5,\)

so

\(z^2+4z-5=0.\)

Factorising,

\((z-1)(z+5)=0.\)

Since \(z=\mathrm e^{3y-4}\gt0\), we reject \(z=-5\), so \(z=1\).

Thus

\(\mathrm e^{3y-4}=1,\)

which gives

\(3y-4=0.\)

Therefore

\(y=\dfrac43.\)