How do you solve #ln x - ln (x-3) = ln 5?#?

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Answer 1 · 2015-07-08T03:05:38

use of logarithm property and then antilog

remember

#lna-lnb=ln(a/b)#

so applying it here we see that

#lnx-ln(x-3)=ln5# can be rewritten as

#ln(x/(x-3))=ln5#

now taking antilog on both sides we get

#antiln(ln(x/(x-3)))=antiln(ln5)#

#x/(x-3)=5#

solving the equation reveals

#x=15/4#

please feel free to comment if you find any mistake
Cheerio!