How do you solve # 2 log 3 + log x = log 36#?

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Answer 1 · 2016-02-07T16:38:36

See solution below.

We can start the solving process by using the log rule #blogn = logn^b#

#log9 + logx - log36 = 0#

Using the property #log_an + log_am = log_a(n xx m)# and #log_an - log_am = log_a(n/m)# we can continue the solving process.

#log((9 xx x)/36)# = 0

Convert to exponential form

#(9x) / 36 = 10^0#

9x = #1 xx 36#

x = 4

Hopefully this helps!