Solve for x. 9 ^(x+2) = 27?

9^(x + 2) = 27
(3^2)^(x + 2) = 3^3
3^[2(x + 2)] = 3^3
2(x + 2) = 3
2*x + 2*2 = 3
2x + 4 = 3
2x = 3 - 4
x = -1/2 (-0.5)

x isn't 1. There are two ways to solve this. I'll begin with the fast way.

You can immediately recognize that 9^(3/2) = 27.

So, x + 2 = 3/2 so x = -1/2.

OR, the second way if you don't recognize that...

9^(x + 2) = 27. Take the logarithm of both sides.

(x + 2)*log(9) = log(27).

x + 2 = log(27)/log(9) = 3/2.

x = -1/2.

Notice that either way you end up with the correct answer.

Hope this helps.

(x + 2) log 9 = log 27
(x + 2) = log 27 / log 9
(x + 2) = log 3 ³ / log 3 ²
x + 2 = 3 log 3 / 2 log 3
x + 2 = 3 / 2
x = - 1/2

9^(x+2) = 27

9 = 3^2 hence,

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

ignore the bases and take the exponents

2(x+2) = 3

get x

x= -1/2

that's it!

9 ^(x+2) = 27

rewrite as:
3^2(x + 2) = 3^3

Now that the bases are both 3, set the exponents equal to each other and solve for x

2(x + 2) = 3
2x + 4 = 3
2x = -1
x = -1/2

Solution: x = -1/2

I hope this helps you
:o)
dr

X= -.5011