perplexus.info :: Just Math : Multiply by 2

Multiply by 2 (Posted on 2021-06-04)

Find all real solutions of

2^x + 3^x - 4^x + 6^x - 9^x = 1

No Solution Yet Submitted by Danish Ahmed Khan

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solution

| Comment 2 of 4 |

Set a=2^x and b=3^x.

Then the equation becomes

a + b - a^2 + ab - b^2 - 1 = 0

a^2 - a(b+1) + b^2 - b + 1 = 0

This is quadratic in a and to stay in the reals the discriminant must be non-negative.

disc = (b + 1)^2 - 4( b^2 - b + 1) >= 0

Simplifying gives -3(b - 1)^2 >= 0 which is satisfied only when b=1.

Then 1 = 3^x and x = 0 is the only solution.

Posted by xdog on 2021-06-05 12:09:08

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