6*4^x-13*6^x+6*9^x=0

Solutions can be seen more easily from the following:

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Integer solutions are x = + or - 1

http://www.wolframalpha.com/input/?i=6*4%5Ex-13*6%5Ex%2B6*9%5Ex%3D0

Thanks Alan

I would not have thought to do that.

How can you tell that there will be another solution as as   $$x\rightarrow -\infty$$    ?

Because, if you look at the original form of the equation, 4^-∞ = 0, 6^-∞ = 0 and 9^-∞ = 0 i.e. each of the terms on the LHS is individually zero.

In that limit my rearranged version of the equation doesn't hold, because a divide by zero would be involved.

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Thanks Alan, that makes sense :)