Answer: \(x=\frac{3+\sqrt6}{2}\), \(u=\frac{4e}{e-2}\), \(v=\frac{13}{5}\).

(a) From

\(\log_6(2x-3)=\frac12,\)

write the equation in exponential form:

\(2x-3=6^{1/2}.\)

So

\(2x-3=\sqrt6.\)

Therefore

\(x=\frac{3+\sqrt6}{2}.\)

(b) Use the logarithm law \(\ln A-\ln B=\ln\frac AB\):

\(\ln\left(\frac{2u}{u-4}\right)=1.\)

Since \(1=\ln e\),

\(\frac{2u}{u-4}=e.\)

Hence

\(2u=e(u-4).\)

So

\(2u=eu-4e.\)

Rearranging,

\(u(2-e)=-4e.\)

Therefore

\(u=\frac{4e}{e-2}.\)

(c) Since \(27=3^3\) and \(9=3^2\),

\(\frac{3^v}{27^{2v-5}}=9\)

becomes

\(\frac{3^v}{(3^3)^{2v-5}}=3^2.\)

So

\(\frac{3^v}{3^{6v-15}}=3^2.\)

Using index laws,

\(3^{v-(6v-15)}=3^2.\)

Thus

\(3^{15-5v}=3^2.\)

Equating powers,

\(15-5v=2.\)

Therefore

\(v=\frac{13}{5}.\)