Solving log equations: log_2(x - 2) = 2 - log_2(x - 5)

[amanda_burnett]

solve for x

log[sub:tfdltgst]2[/sub:tfdltgst](x-2) = 2 - log[sub:tfdltgst]2[/sub:tfdltgst] (x-5)

 

[Deleted member 4993]

Re: Solving a log

solve for x

log[sub:1adrg7sq]2[/sub:1adrg7sq](x-2) = 2 - log[sub:1adrg7sq]2[/sub:1adrg7sq] (x-5)

Please show us your work/thoughts - indicating where you are stuck - so that we know where to begin to help you.

Hint:

Isolate the log terms to one side of the "equality" sign and apply laws of logs.

log[sub:1adrg7sq]2[/sub:1adrg7sq](x-2) + log[sub:1adrg7sq]2[/sub:1adrg7sq] (x-5) = 2

Now continue....

 

[amanda_burnett]

Re: Solving a log

I don't know how to isolate the x

 

[Deleted member 4993]

Re: Solving a log

I don't know how to isolate the x

I have isolated the 'x' for you - now apply laws of Logs.

 

[amanda_burnett]

Re: Solving a log

would it be log[sub:2tukgmp2]2[/sub:2tukgmp2](x[sup:2tukgmp2]2[/sup:2tukgmp2] - 7x + 10)

 

[Deleted member 4993]

Re: Solving a log

would it be

Yes

log[sub:3ezo28um]2[/sub:3ezo28um](x[sup:3ezo28um]2[/sup:3ezo28um] - 7x + 10) = 2

And continue...

 

[amanda_burnett]

Re: Solving a log

so the solution is
x= 1
or x=6

I don't know why I didn't get it in the first place
THANKS : )

 

[Deleted member 4993]

Re: Solving a log

so the solution is
x= 1 <<< this is an extraneous solution. The original equation consist of functions which does not exist at x=1.
or x=6

good work

I don't know why I didn't get it in the first place
THANKS : )