How do you write #5^x=25# in log form?

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Answer 1 · 2018-03-16T18:16:44

I tried this:

I would use the definition of log:
#log_ba=x#
so that: #b^x=a#

In our case we have:
#5^x=25#
so that we get:
#log_5(25)=x#

By the way: #x=2#

Answer 2 · 2018-03-16T18:17:17

#log_5##25=x#

#5^x=25#
#5^x=5^2#
#x=2# (By comparing)