Question 669212
<pre>
3<sup>x+1</sup> + 3<sup>2x+1</sup> = 270

Use the principle that {{{A^(B+C) = A^B*A^C}}} on each term on the left:

3<sup>x</sup>·3<sup>1</sup> + 3<sup>2x</sup>·3<sup>1</sup> = 270

3<sup>x</sup>·3 + 3<sup>2x</sup>·3 = 270

Divide through by 3

3<sup>x</sup> + 3<sup>2x</sup> = 90

Rearrange the equation with 0 on the right:

3<sup>2x</sup> + 3<sup>x</sup> - 90 = 0

Let u = 3<sup>x</sup>
Then u² = 3<sup>2x</sup>

u² + u - 90 = 0

(u - 9)(u + 10) = 0

u - 9 = 0;   u + 10 = 0
    u = 9         u = -10

Using u = 9

Since u = 3<sup>x</sup>

3<sup>x</sup> = 9

Write 9 as 3²

3<sup>x</sup> = 3²

Since the base 3 is the same on both
sides, is positive and not 1, we can 
equate the exponents:

x = 2

Using u = -10
Since u = 3<sup>x</sup>

3<sup>x</sup> = -10

Since all powers of 3 are positive,
and -10 is negative, there is no
solution to this part, and so u=-10
is extraneous.

Therefore x=2 is the only solution.

Edwin</pre>