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Re: If 2^x = 5 and 4^y = 20, what is the value [#permalink]
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Even though an explanation is given, I just wanted to post mine to see if it's also a correct method and maybe someone else had a similar train of thought.

\(2^x = 5\) and \(4^y = 20\)

\(2^x = 5\) and \(2^2^y = 2^2 * 5\)

Here we recognize that \(2^x = 5\) and substitute it into the second equation.

\(2^2^y = 2^2 * 2^x\)

\(2^2^y = 2^x ^+ ^2\)

\(2y = x+2\)

\(2y - 2 = x\)

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Answer : C
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If 2^x = 5 and 4^y = 20, what is the value [#permalink]
\(4^y=20\)

\(4^y = 4 \times 5\)

Since we know that \(2^x = 5\)

\(4^y = 4 \times 2^x\)

Now, since the variables are in the exponent, lets equalize the bases so that we can compare the exponents

\((2^2)^y = 2^2 \times 2^x\)

\(2^{2y} = 2^{2+x}\)

Since the base on both sides of the equation is \(2\), we can equate the exponents

\(2y = 2 + x\)

\(x=2y-2\)

The answer is C.
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