How do you write the equation #log_(1/5) (25)=-2# in exponential form?

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Answer 1 · 2016-10-08T06:16:51

#log_(1/5)25 = -2 " " hArr " " (1/5)^-2 = 25#

Recall: #" "log_a b = c " "hArr " " a^c = b#

The two forms are interchangeable and give the same information, just with a different subject.

Remember: "The base stays the base and the other two swop around"

#log_10 100 = 2" "# asks the question:

"What power of 10 will give 100?" Or

"What index will make 10 into 100?" The answer is 2.

#10^2 = 100" "# states that 10 raised to the power of 2 will give the answer 100.

#log_(1/5)25 = -2 " " hArr " " (1/5)^-2 = 25#

Let's check:

#(1/5)^-2 = (5/1)^2 =25#