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How do you solve #log_10 (2^x)=log_10 32#?

1 Answer

Nov 13, 2016

#x=5#

Explanation:

Rewrite #32# as a productof #2#

#32=2*2*2*2*2=2^5#

#:.log_10(2^x)=log_10(2^5)#

#2^x=2^5#

#x=5#

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