Question

Find all values of c such that 3^2c+1=28*3^c-9. If you find more than one value of c, then list your values in increasing order​

Answer 1

$3^{2c} + 1 = 28\times3^c - 9 \implies 3^{2c} - 28\times3^c = -10$

Complete the square on the left side:

$3^{2c} - 28\times3^c = \left(3^{2c}-28\times3^c+14^2\right)-14^2 = \left(3^c-14\right)^2 - 196$

Then the equation becomes

$\left(3^c-14\right)^2 - 196 = -10 \\\\ \left(3^c-14\right)^2 = 186 \\\\ 3^c - 14 = \pm\sqrt{186} \\\\ 3^c = 14\pm\sqrt{186}$

Both 14 + √186 and 14 - √186 are positive numbers, so we can take the logarithm (base 3) of both sides without issue:

$\log_3\left(3^c) = c = \log_3\left(14\pm\sqrt{186}\right)$

Then in increasing order, the solutions are

c = log₃(14 - √186), c = log₃(14 + √186)