2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

Question

(A) 18

(B) 20

(C) 21

(D) 22

(E) 24

BrentGMATPrepNow · Answer

Let's do some factoring...

Given: 2^x + 2^(x + 3) = (6^2)(2^18)
Factor left side: 2^x(1 + 2^3) = (6^2)(2^18)
Rewrite 6^2 as (3^2)(2^2) to get: 2^x(1 + 8) = (3^2)(2^2)(2^18)
Simplify both sides: (2^x)(9) = (3^2)(2^20)
Rewrite 9 as 3^2 to get: (2^x)(3^2) = (3^2)(2^20)

We can see that x = 20
Answer: B

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Abhishek009 · Answer

2^x + 2^{x + 3} = 6^2*2^{18}

Or, 2^x ( 1 + 2^3 ) = 2^2*3^2*2^{18}

Or, 2^x *3^2 = 3^2*2^{20}

Or, 2^x = 2^{20}

Hence, x = 20
  
Answer will be (B) 20

Nishantg779 · Answer

How did we get 3^2 on the LHS ?

ScottTargetTestPrep · Answer

Simplifying, we have:

2^x + (2^x)(2^3) = 6^2 * 2^18

Factoring out the common 2^x from both terms on the left side of the equation, we have:

(2^x)(1 + 2^3) = 36 * 2^18

(2^x)(9) = 36 * 2^18

2^x = 4 * 2^18

2^x = 2^20

x = 20

Answer: B

CrackverbalGMAT · Answer

In questions on exponents, always apply the rules of exponents to break down complex powers, as far as you can. Also, when there are constants, prime factorise the bases if the bases are composite numbers.

In a question like this, when you see a 6 on the RHS, you might wonder “ How will I get a 3 on the LHS because the LHS seems to be only powers of 2?”. Just go ahead and apply the rules of exponents and simplify the expression on the LHS, you will realise that there’s actually a multiple of 3 on the LHS.

LHS =  2^x  +  2^{x+3}

X^{m+n} = X^m * X^n  .

Applying this in the expression above, LHS =  2^x + 2^x * 2^3 = 2^x   = 2^x  = 2^x   .

RHS =  6^2 * 2^{18} = 2^2* 3^2* 2^{18} = 2^{20} * 3^2 .

You see that both the LHS and the RHS have a 9, which can be cancelled out. On cancelling out the 9, we will be left with  2^x = 2^{20} . Since these are equal numbers with the same bases, the exponents also have to be equal. x = 20.
The correct answer option is B.

Hope that helps!

CrackverbalGMAT · Answer

Just break the equation by prime factorizing. 
 
6^2 = (2^2)*(3^2)

RHS now becomes (2^20)*(3^2)

Equate RHS & LHS now. It will look like:
 
2^x + 2^(x+3) = (2^20)*(3^2) ---------> Eqn 1

Here, LHS does not look similar to RHS. Let's try making them look similar. 
 
2^x + 2^(x+3) can be written as 2^x + (2^x)*(2^3)
Notice 2^x is common in both the terms

Thus, 2^x + (2^x)*(2^3) -> (2^x)*(1+2^3)

OR (2^x)*(9) since 2^3 = 8

Now, go back to Eqn 1, it will look like:
 
(2^x)*(9) = (2^20)*(3^2)

OR
 
(2^x)*(9) = (2^20)*(9) since 3^2 = 9

put x = 20, both equations become just the same. That's what we want.

bumpbot · Answer

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2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

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2^x + 2^(x + 3) = (6^2)(2^18). What is the value of x?

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