Algebra equations

Question

Please help me with these equations! I've been trying all day with no luck.

 

Isolate x in each equation:

 

1) 9^2x = 2(9^x) + 3

2) 3(3^x) + 9(3^-x) = 28

Answer 1

#2.  3(3^x) + 9(3^-x) = 28,  becomes:

 

     3(3^x ) + 9/3^x = 28.

 

Multiply each term by 3^x:  3(3^x)(3^x) + 9 = 28(3^x).

 

Simplify and set equal to 0:  3(3^2x) - 28(3^x) + 9 = 0.

 

Factor:  (3(3^x) - 1)(3^x - 9)=0.

 

3(3^x) - 1 = 0      and 3^x - 9 = 0

 

3(3^x) = 1       and   3^x = 1/3   therefore, x = -1

 

3^x = 9      and   therefore x = 2

 

Check to verify.

Answer 2

Try using the law of  Logs or natural log (ln)

 

#1 . I solved for x which is not what you were asking:  9^2x -2(9^x)-3=0 factors into (9^x-3)(9^x +1)  : 

 

                                                                                                            9^x=3; x=1/2  The other factor is Imaginary