How do you solve #log y = log (x-1) + 1#?

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How do you solve #log y = log (x-1) + 1#?

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Answer 1 · 2016-03-25T10:20:50

y = #10 (x-1)#, x > 1.

Explanation:

Working on common logarithm, #y = 10^((log(x-1)+1)#.
#10^(m+n)= 10^m 10^n# and #10^(log a)=a#.
#y = (x-1) X 10^1=10(x-1)#
Here, #log(x-1)# is defined for x > 1. So, the answer is suject yo x > 1..

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How do you solve #log y = log (x-1) + 1#?

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