Solve 2^4x-9-4x+4*3^2-2x+x-9+x

$$2 ^ { 4 x - 9 - 4 x + 4 } \cdot 3 ^ { 2 - 2 x + x - 9 + x }$$

Evaluate

$$\frac{1}{69984}\approx 0.000014289$$

Solution Steps

Combine 4x and -4x to get 0.

$$2^{-9+4}\times 3^{2-2x+x-9+x}$$

Add -9 and 4 to get -5.

$$2^{-5}\times 3^{2-2x+x-9+x}$$

Calculate 2 to the power of -5 and get \frac{1}{32}.

$$\frac{1}{32}\times 3^{2-2x+x-9+x}$$

Combine -2x and x to get -x.

$$\frac{1}{32}\times 3^{2-x-9+x}$$

Subtract 9 from 2 to get -7.

$$\frac{1}{32}\times 3^{-7-x+x}$$

Combine -x and x to get 0.

$$\frac{1}{32}\times 3^{-7}$$

Calculate 3 to the power of -7 and get \frac{1}{2187}.

$$\frac{1}{32}\times \frac{1}{2187}$$

Multiply \frac{1}{32} and \frac{1}{2187} to get \frac{1}{69984}.

$$\frac{1}{69984}$$

Factor

$$\frac{1}{2 ^ {5} \cdot 3 ^ {7}} = 1.4288980338363053 \times 10^{-5}$$