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avatar+278 

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
avatar+20164 
+3

The second one for today is also about solving

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

 

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

 

laugh

heureka  Nov 1, 2017
 #2
avatar+278 
+3

Heureka,

 

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

juriemagic  Nov 1, 2017
 #3
avatar+91215 
+4

7x^2 - 9  = 1    ...... note  1  =  70

 

 So

 

7x^2 - 9  =  70     since the bases are the same, solve for the exponrents

 

x^2  - 9  = 0

 

(x + 3) (x - 3)  = 0

 

Set each factor to 0   and solve for x  ...so   ....x = 3   or  x = - 3

 

 

cool cool cool

CPhill  Nov 1, 2017

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