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How do you write #a^2 = 36# in log form?

1 Answer

A. S. Adikesavan

May 9, 2016

In this form, #a=+-6#. The logarithmic form is #log_a 36 = 2# and, solving this, a = 6..

Explanation:

If #x=b^y#, the inverse relation is #y=log_b x#.

Here, x = 36, b = a and y = 2.

So, #log_a 36=log_a(6^2)=2 log_a 6=2#.

Now, #log_a 6 =1#. So, a = 6, as #log_a a=1#..

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