+1124 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?..
+26367 The second one for today is also about solving
\(\mathbf{7^{x^2-9}=1}\)
\(\begin{array}{|rcll|} \hline 7^{x^2-9} &=& 1 \quad & | \quad \cdot 7 \\ 7^{x^2-9}\cdot 7 &=& 7 \\ 7^{x^2-9}\cdot 7^1 &=& 7 \quad & | \quad a^b * a^c = a^{b+c} \\ 7^{x^2-9+1} &=& 7 \\ 7^{x^2-8} &=& 7^1 \\ x^2-8 &=& 1 \\ x^2 &=& 9 \quad & | \quad \text{square root both sides} \\ x &=& \pm 3 \\ \hline \end{array}\)
+1124 Heureka,
thank you kindly...something so small that i just do not think of...I do appreciate your help!
