How do you solve #log (x + 9) - log x = 3#?

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Answer 1 · 2016-03-12T15:42:59

Use the log property #log_an - log_am = log_a(n/m)#

#log(x + 9) - log(x) = 3#

#log((x + 9)/(x)) = 3#

Since nothing is noted in subscript, the log is in base 10.

#(x + 9)/x = 10^3#

#(x + 9)/x = 1000#

We can now solve as a simple linear equation by using the property #a/b = m/n => a xx n = b xx m#

#x + 9 = 1000x#

#9 = 999x#

#9/999 = x#

#1/111 = x#

Hopefully this helps!