Question

How do you solve `36^(x^2+9) = 216^(x^2+9)`?

Answer 1

`x=+-sqrt((2kpii)/ln 6-9)`

for any integer `k`

Explanation:

Given:

`36^(x^2+9) = 216^(x^2+9)`

Divide both sides by `36^(x^2+9)` to get:

`1 = 6^(x^2+9) = e^((x^2+9)ln 6)`

From Euler's identity we can deduce:

`(x^2+9)ln 6 = 2kpii`

for any integer `k`

So:

`x^2+9 = (2kpii)/ln 6`

Hence:

`x = +-sqrt((2kpii)/ln 6-9)`

If `k=0` that gives us:

`x = +-sqrt(-9) = +-3i`

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Footnote

If you would like the other roots in `a+bi` form, then you can use the formula derived in https://socratic.org/s/aEUsUcjD , namely that the square roots of `a+bi` are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)`