SOLUTION: I don't understand how to solve a certain type of exponential function. The textbook is not explaining the full steps. The textbook example: line 1 {{{ 3(3^2x) - 28(3^x) + 9

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Question 1098789: I don't understand how to solve a certain type of exponential function. The textbook is not explaining the full steps. The textbook example:

line 1

line2

line 3


line 4

line 5

From here the equation gets factored down until the final answer is x= -1 or x = 2. I understand that part and I follow the first three lines. I get that z is substituted for

I don't follow what is happening in lines 4 and 5. Could someone please explain in detail those two lines?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
Check your line 1.    


I think that it must be 


 = 0.   (In plain text format form  it is  3*(3^(2x)) - 28*3^x + 9 = 0.)


If it is so, then it is a quadratic equation relative .


For many solved problems of this type (your samples) see the lesson
    - Solving exponential equations
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Exponential equations".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


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Comment from student: No. Equation is written as in the textbook. It is a lesson not on quadratic equations but on exponential equations. Thank you for the link.
-------------


My response. Below is the plot of the function in the lft side of your 1-st line.




Plot y = 3(3^2x) - 28(3^x) + 9 =



There is NO ANY SENSE to argue with me.



Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

You don't show the first two lines correctly in your message to us... but apparently you understand what they are supposed to say.

You show line 1 as

;
it should be


And you show line 2 as

; it should be


Then you understand that a substitution was made to help avoid some possible confusion, leaving us with the equation

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....

With this method of factoring, you multiply the leading coefficient and the constant...; then you look for two numbers whose product is that number 27 and whose sum is -28, the coefficient of the linear term.

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:


Now you factor by grouping. You factor a common factor of 3z out of the first two terms: , and you factor a common factor of -1 out of the last two terms:

That gives you your line 5:

And finally you factor out the common linear factor (z-9) to get


(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 to by some other factoring method that you know.)

Now you finish solving this for z: z = 1/3 or z = 9.

Then replacing z with 3^x and solving, you get your two answers:
--> x = -1; and
--> x = 2
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