How do you find the inverse of #f(x) = log7^x#?

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Answer 1 · 2015-12-31T00:59:10

#f^-1(x)=x/log7#

Write as #y=log(7^x)#.

Switch the #x# and #y#, then solve for #y#.

#x=log(7^y)#

Rewrite using logarithm rules.

#x=ylog7#

#y=x/log7#

This can be written in function notation:

#f^-1(x)=x/log7#

These graphs should be reflections over the line #x=y#.

graph{(y-log(7^x))(y-x/log7)=0 [-18.02, 18.02, -9.01, 9.01]}