How to solve for x?

Question

3^2x+7=27^x

Answer 1

Using Susan's method:

 

3^2x+7 = 27^x

3^2x+7 = (3³)^x

3^2x+7 = 3^3x

2x+7 = 3x

7 = x

Answer 2

Dear Youssef,

 

     There is another way to do this problem.  If bases are the same, then you can make the powers equal.

 

Here is an example like yours:  2 ^4X+8 = 16 ^2X .  Sixteen can be written with a base of 2.

 

2 ^(4x+8) = (2⁴ )^2X          2 ^(4X+8) =2 ^8X    The bases are equal, so the exponents must be equal.         

                                                             

                                                              Therefore, 4X+8=8X

 

                                                                  8=8X-4X           8=4X                  X=2

 

Then you can check the answer by substituting for "X."

 

 

Does 2 ^(4(2)+8) = 2 ^(4*4) ?                   2 ¹⁶= 2 ¹⁶             True

 

If I helped you, please give me a thumbs up.

 

                                                               Susan C.

Answer 3

1. 3^2x+7 = 27^x

2. log₃3^2x+7 = log₃27^x

3. (2x+7)(log₃3) = (x)(log₃27)

4. (2x+7)(1) = (x)(3)

5. 2x+7 = 3x

6. 7 = x

7. 3²¹ = 27⁷