How do you solve #5^21 times 4^11 = 2 times 10^n#?

3 Answers
Shwetank Mauria
Jun 28, 2016

#n=21#

Explanation:

As #5^21xx4^11=2xx10^n# hence

#21log5+11log4=log2+n#

Hence #n=21log5+11log2^2-log2#

= #21log5+22log2-log2#

= #21log5+21log2#

= #21(log5+log2)#

= #21log10=21#

Eddie
Jun 28, 2016

#n = 21#

Explanation:

#5^21 * 4^11 = 2 * 10^n#

#(10/2)^21 * (2^2)^11 = 2 * 10^n#

#10^{21} * color{red}{(1/2)^21 * (2)^22} = 2 * 10^n#

#10^{21} * 2 = 2 * 10^n#

#n = 21#

Cesareo R.
Jun 28, 2016

#n=21#

Explanation:

#5^{21}cdot 4^{11} = 2 cdot 10^n#
#5^{10}cdot 5^{10}cdot 4^{10} cdot 5 cdot 4 = 2 cdot 10^n#
#(5 cdot 5 cdot 4)^10cdot 5 cdot 4 = 2 cdot 10^n#
#100^10cdot 20 = 2 cdot 10^n#
#10^{20}cdot 20 = 2 cdot 10^n#
#2cdot 10^{21} = 2 cdot 10^n#
#n = 21#

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