exponents: If 5^21 * 4^11 = 2 x 10^n, then what is n?

[dwill]

If 5^21 * 4^11 = 2 x 10^n
then what is the value of n?

I tried multiplying and I got 20^32.
so, n=32

However, the solution has n=21

Please explain.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\left( 4 \right)^{11} = \left( {2^2 } \right)^{11} = 2^{22} \\
\left( {5^{21} } \right)\left( {2^{22} } \right) = 2\left( {5^{21} } \right)\left( {2^{21} } \right) = 2\left( {\left[ 5 \right]\left[ 2 \right]} \right)^{21} = 2\left( {10^{21} } \right) \\
\end{array}\)

 

[dwill]

\(\displaystyle \L
\begin{array}{l}
\left( 4 \right)^{11} = \left( {2^2 } \right)^{11} = 2^{22} \\
\left( {5^{21} } \right)\left( {2^{22} } \right) = 2\left( {5^{21} } \right)\left( {2^{21} } \right) = 2\left( {\left[ 5 \right]\left[ 2 \right]} \right)^{21} = 2\left( {10^{21} } \right) \\
\end{array}\)

Can you expain why when multiplying 5^21 by 2^22, you factored out a 2?
Can you not simply multiply across?

 

[pka]

\(\displaystyle \L 2^{22} = 2^{\left( {21 + 1} \right)} = \left( {2^{21} } \right)\left( {2^1 } \right) = 2\left( {2^{21} } \right)\)

 

[Denis]

Can you not simply multiply across?

No. Non. Nyet. Uh uh.

 

[soroban]

Hello, dwill!

Another approach . . .

If \(\displaystyle 5^{21}\cdot4^{11}\:=\:2\cdot10^n\), find \(\displaystyle n\).

We have: \(\displaystyle \:2\,\cdot\,10^n\:=\:5^{21}\,\cdot\,4^{11}\)

. . The left side is: \(\displaystyle \:2\,\cdot\,10^n\:=\:2\,\cdot\,(2\cdot5)^n\:=\:2\,\cdot\,2^n\,\cdot\,5^n\)

. . The right side is: \(\displaystyle \:5^{21}\,\cdot\,4^{11}\:=\:5^{21}\,\cdot\,(2^2)^{11}\:=\:2^{22}\,\cdot\,5^{21}\)


The equation becomes: \(\displaystyle \:2\,\cdot\,2^n\,\cdot\,5^n\:=\:2^{22}\,\cdot\,5^{21}\)

. . Divide by \(\displaystyle 2:\;\:2^n\,\cdot\,5^n\:=\:2^{21}\,\cdot\,5^{21}\)

. . And we have: \(\displaystyle \:10^n\:=\:10^{21}\)


Therefore: \(\displaystyle \:n\:=\:21\)