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Mathematics 27 Online
OpenStudy (sphott51):

solve lnx - ln6 = 0

OpenStudy (freckles):

you can add ln(6) on both sides

OpenStudy (freckles):

assume a,b>0 ln(a)=ln(b) implies a=b

OpenStudy (sphott51):

wait what? lol

OpenStudy (sphott51):

So like.. Ln6x =0 ?

OpenStudy (freckles):

Have you tried doing what I suggested? Adding ln(6) on both sides?

OpenStudy (freckles):

and then compare the insides

Nnesha (nnesha):

hmm can we apply quotient rule here ?

OpenStudy (freckles):

also a-a=0 for what number x is ln(x) the same as ln(6)

OpenStudy (freckles):

you could @Nnesha

OpenStudy (anonymous):

\[\log(\heartsuit)=\log(\spadesuit)\iff \heartsuit=\spadesuit\]

OpenStudy (sphott51):

lnx - ln6 =lne6 ?

OpenStudy (freckles):

ln(x)-ln(6)=0 is the same as saying ln(x)=ln(6) for what number x is both sides the same

OpenStudy (freckles):

use what I said earlier or what satellite just said that if ln(a)=ln(b) then a=b

OpenStudy (sphott51):

So x=6?

OpenStudy (freckles):

yes!

OpenStudy (freckles):

ln(6)-ln(6) is totally zero

OpenStudy (freckles):

so x=6 definitely satisfies the equation

OpenStudy (sphott51):

Thanks!

Nnesha (nnesha):

let me see if i get 6 hmm quotient rule\[\large\rm ln y - ln x = \ln \frac{ x }{ y}\] so \[\ln \frac{x}{6}=0\] take e both sides \[e^{\ln}( \frac{ x }{ 6 })= e^0\] e an ln cancels out e^0 =1 \[\cancel{e^{\ln}}( \frac{ x }{ 6 })= e^0\] \[\frac{ x }{ 6 }=1 \rightarrow x=6\]

OpenStudy (freckles):

on that one step where you have ln(x/6)=0 you can also recall that ln(1)=0 and then solve the equation x/6=1

Nnesha (nnesha):

cool!

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