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Re: 4^(2x) [#permalink]
I am confused on this one you did:
"Let's try one more value for x. How about x = 1
We get:
Quantity A: 4^(2x) = 4⁴ = 256
Quantity B: 16^x = 16² = 256"
The quantities are still EQUAL"

If x=1, then 4^(2(1))= 4^2= 16
and 16^1=16. How did you get 4^4 and 16^2?
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Re: 4^(2x) [#permalink]
Faye214 wrote:
I am confused on this one you did:
"Let's try one more value for x. How about x = 1
We get:
Quantity A: 4^(2x) = 4⁴ = 256
Quantity B: 16^x = 16² = 256"
The quantities are still EQUAL"

If x=1, then 4^(2(1))= 4^2= 16
and 16^1=16. How did you get 4^4 and 16^2?


I guess he meant x = 2, but might mistakenly typed 1.

Anyway the best way to solve this is using algebra than plugging in
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Re: 4^(2x) [#permalink]
Faye214 wrote:
I am confused on this one you did:
"Let's try one more value for x. How about x = 1
We get:
Quantity A: 4^(2x) = 4⁴ = 256
Quantity B: 16^x = 16² = 256"
The quantities are still EQUAL"

If x=1, then 4^(2(1))= 4^2= 16
and 16^1=16. How did you get 4^4 and 16^2?


Good catch!
I have edited my response accordingly.

Cheers,
Brent
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Re: 4^(2x) [#permalink]
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Re: 4^(2x) [#permalink]
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