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

\(y=\dfrac43.\)

Answer: (a) \(\lg\!\left(\frac{a^5}{1000b^4}\right)\). (b) \(x=4\) or \(x=-\frac45\).

(a) Use the logarithm laws

\(k\lg m=\lg(m^k),\qquad \lg M-\lg N=\lg\left(\dfrac MN\right).\)

Also, since the logarithms are base \(10\),

\(3=\lg1000.\)

Therefore

\(5\lg a-4\lg b-3=\lg a^5-\lg b^4-\lg1000.\)

Combining into a single logarithm gives

\(\lg\left(\dfrac{a^5}{1000b^4}\right).\)

(b) Let

\(u=\log_5(x+1).\)

Then

\(\log_{x+1}5=\dfrac1u.\)

The equation becomes

\(u-\dfrac1u=0.\)

Multiplying by \(u\),

\(u^2-1=0.\)

Thus

\(u=1\quad\text{or}\quad u=-1.\)

If \(u=1\), then

\(\log_5(x+1)=1,\)

so \(x+1=5\), giving \(x=4\).

If \(u=-1\), then

\(\log_5(x+1)=-1,\)

so \(x+1=5^{-1}=\dfrac15\), giving \(x=-\dfrac45\).

The logarithms require \(x+1\gt0\) and \(x+1\ne1\). Both values satisfy these conditions, so

\(x=4\quad\text{or}\quad x=-\dfrac45.\)

Answer: (a) \(x=32\) or \(x=\frac12\). (b) \(x=\pm\sqrt{5+\ln5}\).

(a) The logarithms are defined for \(x\gt0\) and \(x\ne1\). Let

\(u=\log_2 x.\)

Then, using the change of base identity,

\(\log_x2=\dfrac{1}{\log_2x}=\dfrac1u.\)

The equation becomes

\(u-4=\dfrac5u.\)

Multiplying by \(u\),

\(u^2-4u=5,\)

so

\(u^2-4u-5=0.\)

Factorising,

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

Thus \(u=5\) or \(u=-1\). Therefore

\(\log_2x=5\quad\text{or}\quad \log_2x=-1,\)

giving

\(x=2^5=32\quad\text{or}\quad x=2^{-1}=\dfrac12.\)

Both values are valid for the logarithms.

(b) Start with

\(\mathrm e^{x^2-3}=25\mathrm e^{7-x^2}.\)

Taking natural logarithms of both sides gives

\(x^2-3=\ln25+7-x^2.\)

Move the \(x^2\) terms to one side and the constants to the other:

\(2x^2=10+\ln25.\)

Since \(\ln25=\ln(5^2)=2\ln5\),

\(x^2=5+\ln5.\)

Therefore

\(x=\pm\sqrt{5+\ln5}.\)