exponential function problem help please

Question

Describe the transformations that must be applied (in correct order) to f(x) = 2^x to obtained the transformed functions g(x) = 3(2)^-2t - 6

Answer 1

2^x = 3(2)^(-2t-6) <--- the other post had parenthesis; yes, it did

 

Takes the log (base 2) of both sides

 

x = log 3(2)^(-2t-6)

 

   = (-2t-6) log3(2)

 

  = (-2t-6) ( log 3 + log 2)

 

 = (-2t-6) ( log 3 + 1)

Answer 2

OK, you've posted this problem several times and haven't gotten it right yet.  I think the problem is:

 

f(x) = 2^x

g(x) = 3·2^2x-6

 

The variable in g(x) needs to be x, not t.  Otherwise it would be g(t).  The general transformations of f(x) = 2^x are:

 

g(x) = a·2^b(x+c) + d

• a = vertical dilation (stretch)
• b = horizontal dilation (stretch)
• c = horizontal shift (c>0 shifts left, c<0 shifts right)
• d = vertical shift (d>0 shifts up, d<0 shifts down)

 

Now put your g(x) into the above form:

 

g(x) = 3·2^2(x-3)

• a = vertical dilation (stretch) = 3
• b = horizontal dilation = 2
• c = horizontal shift to the right = -3
• d = vertical shift = 0

The transformations are applied in the order c, b, then a.