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**Soroban** Hello, ntrantrinh!

Here's #2 . . .

We note that: .$\displaystyle \begin{array}{ccccccc}a & = &\log(11-6\sqrt{2}) &=& (3-\sqrt{2})^2 \\

b &=& \log_7(45 + 29\sqrt{2}) &=& (3+\sqrt{2})^3 \end{array}$

Then: .$\displaystyle 3a+2b \;=\;3\log_7(3-\sqrt{2})^2 + 2\log_7(3-\sqrt{2})^3 $

. . . . . . . . . . $\displaystyle =\;6\log_7(3=\sqrt{2}) + 6\log_7(3+\sqrt{2}) $

. . . . . . . . . . $\displaystyle =\;6\bigg[\log_7(3-\sqrt{2}) + \log_7(3+\sqrt{2})\bigg]$

. . . . . . . . . . $\displaystyle =\;6\,\log_7\!\bigg[(3-\sqrt{2})(3+\sqrt{2})\bigg] $

. . . . . . . . . . $\displaystyle =\;6\,\underbrace{\log_7(7)}_{\text{This is 1}}$

. . . . . . . . . . $\displaystyle =\; 6$