Let's first write \(4^x\) as a power of \(2\) (like we have in the other equation)
Replace \(4\) with \(2^2\) to get: \(4^x = (2^2)^x= 2^{2x}\)
So we now want to find a value that is equivalent to \(2^{2x}\)

Now let's do something with the first equation.
Take: \(t = 2^{x + 1}\)
Rewrite as: \(t = (2^x)(2^1)\)
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Aside: Some students won't like that last step, so here's why it works:
If we take \((2^x)(2^1)\) and find the product by using the product law we get: \(2^{x+1}\)
So as you can see that last step was mathematically legitimate
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Simplify to get: \(t = (2^x)(2)\)
Divide both sides by \(2\) to get: \(\frac{t}{2} = 2^x\)

We're almost there!
We want to find a value that is equivalent to \(2^{2x}\)

If we take: \(\frac{t}{2} = 2^x\)
And square both sides we get: \((\frac{t}{2})^2 = (2^x)^2\)
Simplify to get: \(\frac{t^2}{4}= 2^{2x}\)

Answer: E

Cheers,
Brent
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\(t = 2^{x + 1}\)
so t^2 = 2^2x. 4
t= 4^x*4 /4 ; 4^x
option e

E.

t = 2^(X+1)
Squaring both sides:

t^2 = 2^2(X+1)
or, t^2 = 4^(X+1)
4^(X+1) can be written as 4^X * 4^1
==> t^2 = 4^X * 4
or 4^X = t^2 / 4
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\(t = 2^{x + 1}\)

\(t = 2(2^x)\)

\(\frac{t}{2} = 2^x\)

\((\frac{t}{2})^2 = (2^x)^2\)

\(\frac{t^2}{4} = 2^{2x} = 4^x\)

Answer is E.
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