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Answer 1 · 2017-04-05T08:40:34

Solution to a) #->x=1/2#

Solution to a)

Adding logs reflects the action of multiplication of the source values.

So #ln(x-1)+ln(3)=ln(x)# is the same as: #ln( 3[x-1])=ln(x)#

Using an example: note that #ln^(-1)(a)=a#

So #" "ln^(-1)(3[x-1])" "=" "ln^(-1)(x)#

#" "3(x-1)" "=" "x#

#" "3x-1=x#

#" "2x=1#

#" "x=1/2#

Answer 2 · 2017-04-05T08:54:12

Solution to b) #" "x~~8.773...# to 3 decimal places

Solution to b)

Using an example: note that #" "2ln(a) -> ln(a^2)#

So we have:

#ln[(5x)^3]-ln(x^2)=7#

#ln[(125x^3)/x^2]=7#

#ln(125x)=7#

#ln^(-1)(125x)=ln^(-1)(7) larr" not convinced this will work!"#

Trying something else!

#ln(125)+ln(x)=7#

#ln(x)=7-ln(125)#

#x=ln^(-1)(x)= ln^(-1)[color(white)(2/2)7-ln(125)color(white)(2/2)]#

#x~~8.773...# to 3 decimal places