How do you solve #log_7 12 = x#?

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Answer 1 · 2015-12-29T19:53:40

From #log# definition, we have that #log_ba=c <=>b^c=a#

Thus,

#log_7(12)=x <=> 7^x=12#

Now, we can apply #log# on both sides of the equation.

#log7^x=log12#

Another #log# rule states that #log_ba^n=n*log_ba#, so:

#x*log7=log12#
#x=log12/log7# (both on base 10 or base #e# or whatever base you end up finding convenient! It really does not matter!)

Using a calculator, as these are not exact expressions, you can find the approximate value of

#x~=1.277#