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matthewsmith_89
0akshay0
Can you explain your working out in depth. I am not understanding.

Or Bunuel can you help me.

\(3^m*7^{m-1}\)

\((3*3^{m-1})*7^{m-1}\)

\(3*(3^{m-1}*7^{m-1})\)

\(3*(3*7)^{m-1}\)

\(3*(21^{m-1})\)

Hope it's clear.
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==> From \(3^m7^m^-^1=3(3^m^-^1)(7^m^-^1)=3(3*7)^m^-^1=3(21^m^-^1)\), the answer is C.

Answer: C
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3^m x 7^(m-1) = 3 x 3^(m-1) x 7^(m-1) = 3 x (3 x 7)^(m-1) = 3 x 21^(m-1) Option C
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Hey,

Why are we multiplying the equation with a 3? Are we factoring it out?

Bunuel
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0akshay0
Can you explain your working out in depth. I am not understanding.

Or Bunuel can you help me.

\(3^m*7^{m-1}\)

\((3*3^{m-1})*7^{m-1}\)

\(3*(3^{m-1}*7^{m-1})\)

\(3*(3*7)^{m-1}\)

\(3*(21^{m-1})\)

Hope it's clear.
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mikhailhx1
Hey,

Why are we multiplying the equation with a 3? Are we factoring it out?

Bunuel
matthewsmith_89

Which of the following is equal to \(3^m7^{m-1}\)?

A. \( 3(21^m)\)
B. \( 7(21^m)\)
C. \( 3(21^{m-1})\)
D. \( 7(21^{m-1})\)
E. \( 21^{m-1}\)


0akshay0
Can you explain your working out in depth. I am not understanding.

Or Bunuel can you help me.

\(3^m*7^{m-1}\)

\((3*3^{m-1})*7^{m-1}\)

\(3*(3^{m-1}*7^{m-1})\)

\(3*(3*7)^{m-1}\)

\(3*(21^{m-1})\)

Hope it's clear.

Yes, we are breaking \(3^m\) into \(3 * 3^{(m-1)}\) so that both 3 and 7 have the same exponent (m-1). This allows us to apply the rule \(a^m * b^m = (a*b)^m\).
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