The second one for today is also about solving \(x\)

\(7^{x^2-9}=1\)

I tried this:

Multiply both sides by 7, I get:

\(49^{x^2-9}=7\)

\(7^{2(x^2-9)}=7\)

\(2(x^2-9)=1\)

\(2x^2-18=1\)

then proceed untill

\(x^2=9,5\)

The answer after this is incorrect. I was thinking of something else: After this line:

\(2(x^2-9)=1\), do this:

\(2(x+3)(x-3)=1\)

going further yields also a wrong answer. Please help me?..

juriemagic
Nov 1, 2017

#1**+3 **

**The second one for today is also about solving**

\(\mathbf{7^{x^2-9}=1}\)

**\(\begin{array}{|rcll|} \hline 7^{x^2-9} &=& 1 \quad & | \quad \cdot 7 \\ 7^{x^2-9}\cdot 7 &=& 7 \\ 7^{x^2-9}\cdot 7^1 &=& 7 \quad & | \quad a^b * a^c = a^{b+c} \\ 7^{x^2-9+1} &=& 7 \\ 7^{x^2-8} &=& 7^1 \\ x^2-8 &=& 1 \\ x^2 &=& 9 \quad & | \quad \text{square root both sides} \\ x &=& \pm 3 \\ \hline \end{array}\)**

heureka
Nov 1, 2017

#2**+3 **

Heureka,

thank you kindly...something so small that i just do not think of...I do appreciate your help!

juriemagic
Nov 1, 2017