+0

Solve 'x'

0
356
4
+313

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:

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

Nov 1, 2017

#1
+21191
+3

The second one for today is also about solving

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

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

Nov 1, 2017
#2
+313
+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
+95884
+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

Nov 1, 2017