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Please help with this question:-

$(\sqrt{243}+3)^x+(\sqrt{243}-3)^x=15^x$.
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After applying properties of log and doing all calculation I am x=-1.20 negative value and on above equation, if you put x=1 it will give 234/15 which is not equal to 1,

 

sorry x=0 will not be a solution here I wrongly mentioned it
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$(\frac{√243+3}{15})^x+(\frac{√243-3}{15})^x=1$

$√243+3 = 12.58/15 <1 , √243+3 = 18.58/15 >1$

$(\frac{√243+3}{15})^x$ will be >1 if X >1

$(\frac{√243-3}{15})^x$ will be <1 always

so sum of these two cannot be 1 for any real positive number.

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