How do you find the inverse of #y = 1 + log_10 (x + 2) #?

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Answer 1 · 2015-10-10T08:42:53

#f^-1(x)=(10^x-20)/10#

#1# can be written as #log_10 10#.

#y= log_10 10+log_10 (x+2)#

Using the Product Law which states that #log_b AC= log_b A+ log_b C#,

#y=log_10 (10*[x+2])#
#y=log_10 10x+20#

Since #log_a b=c# can be written as #b=a^c#,

#10x+20=10^y#

Make #x# the subject.
#10x=10^y-20#
#x=(10^y-20)/10#

Change #y# to #x# in your final answer.

#f^-1(x)=(10^x-20)/10#