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:

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

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