How do you solve #log_b121=2#?

1 Answer
r3cebarnett
Nov 5, 2016

#+-11#, see explanation.

Explanation:

if you have #a^b = c#, logarithms go #log_ac = b#. That's just a thing that you need to know, I'm not quite sure how to explain that, but if we apply it to the logarithm in the problem...

#log_b121=2 -> b^2=121#.

And from here you just take the square root of both sides and get #+-11#. It's plus or minus because #11^2# and #(-11)^2# equal 121. (Negative multiplied by a negative equals a positive).

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