Finding real value solutions

[kia]

I am so lost. The question is to determine all real value solutions to the equation 3^(2x+2) - 3^(x+3) - 3^x + 3 = 0. If I could get a push, I would really appreciate it.

 

[soroban]

Hello, kia!

Determine all real solutions: $\displaystyle \;3^{2x+2}\,-\,3^{x+3}\,-\,3^x\,+\,3\;=\; 0$

On a hunch, I tried to factor it (by "grouping") . . .

Factor: $\displaystyle \;3^{x+2}\left(3^x\,-\,3\right)\,-\,\left(3^x\,-\,3\right)\;=\;0$

Factor: $\displaystyle \;\left(3^x\,-\,3\right)\left(3^{x+2}\,-\,1\right)\;=\;0$


So we have: $\displaystyle \,3^x\,-\,3\:=\:0\;\;\Rightarrow\;\;3^x\,=\,3\;\;\Rightarrow\;\;x\,=\,1$

. . . . . . and: $\displaystyle \,3^{x+2}\,-\,1\:=\:0\;\;\Rightarrow\;\;3^{x+2}\,=\,1\,=\,3^0\;\;\Rightarrow\;\;x\,+\,2\:=\:0\;\;\Rightarrow\;\;x\,=\,-2$

 

[kia]

Finding real values

Thank you!

Just one more quick question, though.

At the end, where you have 3(x+2)=3^0, can you just equate the exponents? Because I thought about that, but I assumed that in this case, it was mathematically incorrect. Thanks again!!

 

[stapel]

I assumed that in this case, it was mathematically incorrect.

Why?

Please clarify. Thank you.

Eliz.

 

[kia]

I only made the assumption because I would only be factoring the 3^(x+2) term out of the first two numbers and thought that in order for me to be able to factor it, it would have to be in each term.

When I looked at it again, it made sense to me that it was logical. When I take an approach to a problem that I question, I just automatically assume that it is mathematically incorrect. I'm sorry it took me so long to respond.

Kia