How do you solve # log(x-5) + log(x-2)=1#?

2 Answers
Nov 29, 2015

Answer:

#x=7#

Explanation:

#log(x-5)+log(x-2)=1#
#color(white)("XXXXXXXX")#since #log(a*b) = log(a)+log(b)#
#rArrlog( (x-5)(x-2) ) =1#
#color(white)("XXXXXXXX")#expanding the multiplication
#log( x^2-7x+10 ) =1#
#color(white)("XXXXXXXX")#despite other comments the default base for #log# is #10#
#color(white)("XXXXXXXX")#so using the above as exponents of #10#
#10^(log(2x^2-7x+10))=10^1#
#rArrx^2 - 7x +10 = 10#
#color(white)("XXXXXXXX")#subtracting #10# from both sides
#x^2-7x=0#
#color(white)("XXXXXXXX")#factoring
#x(x-7)=0#

i.e. #x=0# or #x=7#

However #log(x-5)# and #log(x-2)# are undefined if #x=0#;
so #x=0# is an extraneous solution
and only #x=7# is valid

Sep 3, 2017

Answer:

#x=7#

Explanation:

As no base is given, it is assumed to be #10#
Natural logs are commonly denoted by ln.

#log(x-5) +log(x-2) =color(blue)(1)#

In an expression or equation, the terms must all be in the same form - either all logs or all numbers.

#log(x-5) +log(x-2) =color(blue)(log10)" "larr(color(blue)(log_10 10 hArr 1))#

#color(white)(xxxxxxxxxxx)#Apply the law: #" "loga + logb hArr log(ab)#

#log((x-5)(x-2)) =log10#

#color(white)(xxxxxxxxxxx)#Apply the law: #" "log a = logb hArr a=b#

#:.(x-5)(x-2) = 10color(white)(xxxxxxxxxxx)#'drop' the logs

#x^2-7x+10 = 10" "larr# solve the quadratic equation

#x^2-7x=0" "larr# factorise

#x(x-7)=0#

#x =0 or x=7#

However, #x=0# is an extraneous solution, and not valid in this equation.

#x=7#