If $\mathrm{x}=\frac{\log \log \mathrm{a}-\log \log \mathrm{b}}{\log \mathrm{a}-\mathrm{blog} \mathrm{b}}$ satisfies the equation

$2^{3^{\mathrm{x}}}=3^{4^{\mathrm{x}}}$

Then a+b=_______.

Taking the logarithm of both sides and using the identity

${\log }_{\mathrm{a}}\left({\mathrm{b}}^{\mathrm{c}}\right)={\mathrm{clog}}_{\mathrm{a}}\mathrm{b}$

gives us

$3^{\mathrm{x}}\log 2=4^{\mathrm{x}}\log 3$

Taking the logarithm of both sides and using the identity

${\log }_{\mathrm{a}}\left(\mathrm{bc}\right)={\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{a}}\mathrm{c}$

along with the previous one, we have

$\mathrm{x} \log 3+\log \log 2=\mathrm{x} \log 4+\log \log 3$

Rearranging terms and noting $4=2 \log 2$, the solution is thus

$\mathrm{x}=\frac{\log \log 3-\log \log 2}{\log 3-2 \log 2}$