# Question #b397b

##### 2 Answers

#### Explanation:

Start by rewriting your equation

#2^(2x) - 5 * 4^(x+1) + 38 = 0#

as

#2^(2x) - 5 * 4^x * 4 + 38 = 0#

As you know, you have

#4 = 2^2#

This implies that

#4^x = (2^2)^x = 2^(2 * x) = 2^(2x)#

This means that the equation can be written as

#2^(2x) - 5* 4 * 2^(2x) + 38 = 0#

At this point, you can take

#2^(2x) * (1 - 5 * 4) + 38 = 0#

This is equivalent to

#2^(2x) = (- 38)/(-19)#

#2^(2x) = 2#

Since

#2^(2x) = 2^1#

This implies that

#2x = 1#

which gets you

#x = 1/2#

To double-check your calculations, plug

#2^((2 * 1/2)) - 5 * 4^((1/2 + 1)) + 38 = 0#

#2^1 - 5 * 4^(3/2) + 38 = 0#

Since

#4^(3/2) = sqrt(4^3) = 4sqrt(4) = 4 * 2 = 8#

you will have

#2 - 5 * 8 + 38 = 0#

#2 - 40 + 38 = 0 " "color(darkgreen)(sqrt())#

#### Explanation:

Note first that

#2^(2x)-5(4^(x+1))+38=0#

#2^(2x)-20(4^x)+38=0#

Rewrite the exponential function with base

#2^(2x)-20(2^(2x))+38=0#

Now, note that

#-19(2^(2x))+38=0#

To solve for

#-19(2^(2x))=-38#

Dividing by

#2^(2x)=2#

#2^color(blue)(2x)=2^color(blue)1#

Since the bases of the exponential functions are equal, so must their exponents:

#2x=1#

#x=1/2#