Question 77

If $$\left(\sqrt{\frac{3}{5}}\right)^a = \left(\sqrt{\frac{625}{81}}\right)^{\frac{a + 3}{2}}$$, then a =


The RHS of the equation can be modified as $$\left(\sqrt{\ \frac{81}{625}}\right)^{-\frac{\left(a+3\right)}{2}}=\left(\sqrt{\ \frac{3}{5}}\right)^{4\cdot\left(-\frac{\left(a+3\right)}{2}\right)}$$

As the bases are equal on both sides, respective powers can be equated:
a = -2*(a+3)
a = -2

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