4^(x+2) = 32     solve for x value

Expert Answers
hala718 eNotes educator| Certified Educator

4^(x+2) = 32

Let us rewrite :

4= 2^2

32= 2^5

Now substitute:

(2^2)^(x+2) = 2^5

==> 2^2(x+2) = 2^5

Then :

2(x+2) = 5

==> 2x + 4= 5

==> 2x= 1

==> x= 1/2

To check :

4^(1/2 + 2) = 32

2^2(5/2) = 32

2^5 = 32

32= 32

 

tonys538 | Student

The equation 4^(x+2) = 32 has to be solved.

To solve this equation either the base or the exponent should be equal.

Here, it can be seen that both 4 and 32 are powers of 2. Rewrite the equation as terms with 2 being raised to a power.

4^(x+ 2)

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

= 2^(2*(x+2))

= 2^(2x + 4)

32 = 2^5

This gives:

2^(2x +4) = 2^5

As the base is the same, equate the exponent

2x + 4 = 5

2x = 1

x = 1/2

The solution of the equation is x = 1/2

giorgiana1976 | Student

This is an exponential equation and we'll solve it use the one to one rule.

Let's see how.

First, we notice that 4 and 32  are multiples of 2.

So, the term 4^(x+2) = 2^2(x+2).

Also we could write 32 = 2^5

Now we'll rewrite the equation:

2^2(x+2) = 2^5

Now we'll use the rule:

2(x+2) = 5

We'll open the brackets:

2x + 4 = 5

We'll add -4 both sides:

2x = 1

We'll divide by 2:

x = 1/2

krishna-agrawala | Student

Given:

4^(x + 2) = 32

We know 4 = 2^2

Therefore:

4^(x + 2) = (2^2)^(x + 2) = 2^[2*(x + 2)]

Therefore:

4^(x + 2) = 2^[2*(x + 2)] = 32

We know 2^5 = 32

Therefore:

2*(x + 2) = 5

2x + 4 = 5

2x = 5 - 4 = 1

x = 1/2

neela | Student

To solve 4^(x+2) = 32.

Solution:

T0 solve the equation , we keep the both sides in powers of the same base.

The left side , here , x+2 to the power of 4. The base is 4. The right side is 32. but 32 = 4*4*4^(1/2) = 4^(2+1/2) = ,as a^m+a^n = a^(m+n). Therefore the given equation becomes:

4^(x+2) = 4^(2+1/2). Since the bases are same on both sides , the power x+2 on the left is equal to the power 2+1/2 on the right.

x+2 =2+1/2

x=1/2.

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