Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

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#

Related questions

See all questions in Logarithmic Models

Impact of this question

982 views around the world

You can reuse this answer
Creative Commons License