How do you prove that #log 12 = log 3 + log 4# ?

2 Answers
George C.
Dec 15, 2017

See explanation...

Explanation:

Note that as a real valued function of real numbers #10^x# is one to one from #(-oo, oo)# onto #(0, oo)#, with inverse #log x# which is one to one from #(0, oo)# onto #(-oo, oo)#.

Now for any real values of #a#, #b#, we have:

#10^(a+b) = 10^a * 10^b#

So:

#log 12 = log (3 * 4)#

#color(white)(log 12) = log (10^(log 3) * 10^(log 4))#

#color(white)(log 12) = log (10^(log 3 + log 4))#

#color(white)(log 12) = log 3 + log 4#

Sunayan S
Dec 15, 2017

Below referred....

Explanation:

Let's ,

#log_(10)12=x#
#=>10^x=12" "......(1)#[all base is 10]

Next, let's

#log_(10)3=m#
#=>10^m=3" ".......(2)#

and,

#log_(10)4=n#
#=>10^n=4" "........(3)#

From #(1),(2),(3)#,

#12=3×4#
#=>10^x=10^m×10^n#
#=>10^x=10^(m+n)#
#=>x=m+n#
#=>log12=log3+log4# (proved)

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