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

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