How do you solve # log (x+7)=1#?

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Answer 1 · 2018-05-09T14:35:13

I tried this:

It depends upon the 'base" of your log. Let us say that is #10#; we can write:

#log_(10)(x+7)=1#

using the definition of log we can write:

#x+7=10^1#

and:

#x=10-7=3#

if you have a different base use the one you got instead of #10#.