How do you solve #log x - log (9x+4)= -1#?

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Answer 1 · 2016-02-27T03:45:24

Use the rule #log_an - log_am = log_a(n/m)# to simplify

#log((x)/(9x + 4)) = -1#

Convert to exponential form. The log is in base 10, since nothing is noted in subscript beside the log.

#x/(9x + 4) = 10^-1#

#x/(9x + 4) = 1/10^1#

#x/(9x + 4) = 1/10#

Simplify using the rule #a/b = m/n -> a xx n = b xx m#

#x(10) = (9x + 4)1#

#10x = 9x + 4#

#10x - 9x = 4#

#x = 4#