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M=4^x+4^x+4^x+4^x
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02 Jun 2024, 23:48
02 Jun 2024, 23:48
Expert Reply
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Question Stats:
88% (01:19) correct
11% (01:57) wrong based on 26 sessions
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\(M=4^x+4^x+4^x+4^x\). What is the value of \(M^2\) ?
A. \(2^{2x+4}\)
B. \(2^{4x+4}\)
C. \(4^{x+4}\)
D. \(4^{2x+4}\)
E. \(4^{4x}\)
A. \(2^{2x+4}\)
B. \(2^{4x+4}\)
C. \(4^{x+4}\)
D. \(4^{2x+4}\)
E. \(4^{4x}\)
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Free Materials for the GRE General Exam - Where to get it!!
Shorter GRE - Preparing for the Quantitative Reasoning Measure
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Free Materials for the GRE General Exam - Where to get it!!
Shorter GRE - Preparing for the Quantitative Reasoning Measure
Shorter GRE - Preparing for the Verbal Reasoning Measure
Re: M=4^x+4^x+4^x+4^x
[#permalink]
06 Dec 2024, 03:51
06 Dec 2024, 03:51
Expert Reply
We know $\(M=4^x+4^x+4^x+4^x\)$; we need to find the value of $\(M^2\)$.
$$
\(M=4^x+4^x+4^x+4^x=4^x(1+1+1+1)=4^x \times 4^1=4^{x+1}\left(\because x^{a+b}=x^a \times x^b\right)\)
$$
So, we get
$\(\mathrm{M}^2=\left(4^{x+1}\right)^2=4^{2(x+1)}=4^{2 x+2}=\left(2^2\right)^{2 x+2}=2^{2(2 x+2)}=2^{(4 x+4)}\left(\because\left(x^a\right)^b=x^{a \times b}=x^{a b}\right)\)$
Hence the answer is (B).
$$
\(M=4^x+4^x+4^x+4^x=4^x(1+1+1+1)=4^x \times 4^1=4^{x+1}\left(\because x^{a+b}=x^a \times x^b\right)\)
$$
So, we get
$\(\mathrm{M}^2=\left(4^{x+1}\right)^2=4^{2(x+1)}=4^{2 x+2}=\left(2^2\right)^{2 x+2}=2^{2(2 x+2)}=2^{(4 x+4)}\left(\because\left(x^a\right)^b=x^{a \times b}=x^{a b}\right)\)$
Hence the answer is (B).
gmatclubot
Re: M=4^x+4^x+4^x+4^x [#permalink]
06 Dec 2024, 03:51
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