How do you solve $36^{x^{2} + 9} = 216^{x^{2} + 9}$ $36^(x^2+9) = 216^(x^2+9)$ ?

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How do you solve $36^{x^{2} + 9} = 216^{x^{2} + 9}$ $36^(x^2+9) = 216^(x^2+9)$ ?

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Answer 1 · 2017-05-22T19:07:25

$x = \pm \sqrt{\frac{2 k π i}{ln 6} - 9}$ $x=+-sqrt((2kpii)/ln 6-9)$

for any integer $k$ $k$

Explanation:

Given:

$36^{x^{2} + 9} = 216^{x^{2} + 9}$ $36^(x^2+9) = 216^(x^2+9)$

Divide both sides by $36^{x^{2} + 9}$ $36^(x^2+9)$ to get:

$1 = 6^{x^{2} + 9} = e^{(x^{2} + 9) ln 6}$ $1 = 6^(x^2+9) = e^((x^2+9)ln 6)$

From Euler's identity we can deduce:

$(x^{2} + 9) ln 6 = 2 k π i$ $(x^2+9)ln 6 = 2kpii$

for any integer $k$ $k$

So:

$x^{2} + 9 = \frac{2 k π i}{ln 6}$ $x^2+9 = (2kpii)/ln 6$

Hence:

$x = \pm \sqrt{\frac{2 k π i}{ln 6} - 9}$ $x = +-sqrt((2kpii)/ln 6-9)$

If $k = 0$ $k=0$ that gives us:

$x = \pm \sqrt{- 9} = \pm 3 i$ $x = +-sqrt(-9) = +-3i$

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Footnote

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

$\pm ⎛ ⎝ ⎛ ⎝ \sqrt{\frac{\sqrt{a^{2} + b^{2}} + a}{2}} ⎞ ⎠ + ⎛ ⎝ \frac{b}{| b |} \sqrt{\frac{\sqrt{a^{2} + b^{2}} - a}{2}} ⎞ ⎠ i ⎞ ⎠$ $+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)$

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How do you solve $36^{x^{2} + 9} = 216^{x^{2} + 9}$ $36^(x^2+9) = 216^(x^2+9)$ ?

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Explanation:

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How do you solve 36^(x^2+9) = 216^(x^2+9)36x2+9=216x2+936^(x^2+9) = 216^(x^2+9)?

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Explanation:

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How do you solve $36^{x^{2} + 9} = 216^{x^{2} + 9}$ $36^(x^2+9) = 216^(x^2+9)$ ?