Logical Log... #1

We know from the first three equations that $\log_a27 = \log_37, \log_b49 = \log_711$ , and $\log_c\sqrt{11} = \log_{11}25$ . Substituting, we find

$a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}$ .

We know that $x^{\log_xy} =y$ , so we find

$27^{\log_37} + 49^{\log_711} + \sqrt {11}^{\log_{11}25}$

$(3^{\log_37})^3 + (7^{\log_711})^2 + ({11^{\log_{11}25}})^{1/2}$ .

The $3$ and the $\log_37$ cancel out to make $7$ , and we can do this for the other two terms. We obtain

$7^3 + 11^2 + 25^{1/2}$

$= 343 + 121 + 5= \boxed {469}$ .