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

1 Answer
George C.
May 22, 2017

#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#

#color(white)()#
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)#

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