Question 1098789
<br>You don't show the first two lines correctly in your message to us... but apparently you understand what they are supposed to say.<br>
You show line 1 as<br>
{{{ 3(3^2x) - 28(3^x) + 9 = 0 }}};<br> it should be
{{{ 3(3^(2x)) - 28(3^x) + 9 = 0 }}}<br>
And you show line 2 as<br>
{{{ 3((3^x)2) - 28(3^x) + 9=0 }}}; it should be
{{{ 3((3^x)^2) - 28(3^x) + 9=0 }}}<br>
Then you understand that a substitution {{{z=3^x}}} was made to help avoid some possible confusion, leaving us with the equation {{{3z^2-28z+9=0}}}<br>
Understanding that part of the problem is much harder than understanding the rest of the solution.  Apparently you are not familiar with (or don't recognize) this method for factoring a quadratic....<br>
With this method of factoring, you multiply the leading coefficient and the constant...{{{3*9=27}}}; then you look for two numbers whose product is that number 27 and whose sum is -28, the coefficient of the linear term.<br>
Those two numbers are -27 and -1; you then break up the middle term into two terms with those coefficients, to get what you show as line 4:
{{{ 3z^2 - 27z - z + 9 = 0 }}}<br>
Now you factor by grouping.  You factor a common factor of 3z out of the first two terms: {{{3z^2-27z = 3z(z-9)}}}, and you factor a common factor of -1 out of the last two terms: {{{-z+9 = -1(z-9)}}}<br>
That gives you your line 5: {{{3z(z - 9) - 1(z - 9) = 0 }}}<br>
And finally you factor out the common linear factor (z-9) to get
{{{(3z-1)(z-9) = 0}}}<br>
(I suspect that is what the "line 6" from the textbook was that you didn't show us.  They just used a particular factoring method that you didn't recognize.  You probably could have gone straight from {{{3z^2-28z+9=0}}} to {{{(3z-1)(z-9) = 0}}} by some other factoring method that you know.)<br>
Now you finish solving this for z: z = 1/3 or z = 9.<br>
Then replacing z with 3^x and solving, you get your two answers:
{{{3^x = 1/3 = 3^(-1)}}}  -->  x = -1; and
{{{3^x = 9 = 3^2}}}  -->  x = 2