SOLUTION: Trying to help my daughter Log (x-5) = log (x-2)+1

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Trying to help my daughter Log (x-5) = log (x-2)+1      Log On


   

Question 812628: Trying to help my daughter
Log (x-5) = log (x-2)+1
Found 2 solutions by jsmallt9, lwsshak3:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log(x-5) = log(x-2) + 1
To solve logarithmic equations like this one, we start by using algebra and/or properties of logs to transform the equation into one of the following forms:
log(expression) = expression
or
log(expression) = log(expression)
Since the second form is "all-log" and since our equation has a "non-log" term on the right (the "1"), we will aim for the first form. So we need to get the logs on one side and then try to combine them. So we'll start by subtracting log(x-2) from each side:
log (x-5) - log (x-2) = 1
Now we can use the log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29 property to combine them:
log%28%28%28x-5%29%2F%28x-2%29%29%29+=+1
And the equation is now in the first form.

The next stage with the first form is to rewrite the equation in exponential form. Since the implied base is 10 we get:
%28x-5%29%2F%28x-2%29+=+10%5E1
which simplifies to:
%28x-5%29%2F%28x-2%29+=+10

Now that the logs are gone we solve the equation. Multiplying by x-2 (to eliminate the fraction) we get:
x - 5 = 10x - 20
Subtracting x:
-5 = 9x - 20
Adding 20:
15 = 9x
Dividing by 9:
15%2F9+=+x
which reduces to:
5%2F3+=+x

The last part is to check. This is not optional! A check must be made to ensure that all bases and arguments are valid. (Arguments must be positive and bases must be positive and not a 1.) Use the original equation to check:
log(x-5) = log(x-2) + 1
Checking x+=+5%2F3:
log%28%28%285%2F3%29-5%29%29+=+log%28%285%2F3%29-2%29%29+%2B+1
Simplifying:
log%28%285%2F3-15%2F3%29%29+=+log%28%285%2F3-6%2F3%29%29+%2B+1
log%28%28-10%2F3%29%29+=+log%28%28-1%2F3%29%29+%2B+1
We can see that both arguments are negative (i.e. invalid). So we must reject this solution. (If only one argument had been invalid we would still reject the solution.) And since we are rejecting the only "solution" we found, there are no solutions to your original equation!

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Log (x-5) = log (x-2)+1
log(x-5)-log(x-2)=1
log[(x-5)/(x-2)]=1
convert to exponential form: base(10) raised to log of the number(1)=number[(x-5)/(x-2)]
10^1=(x-5)/(x-2)
10(x-2)=x-5
10x-20=x-5
9x=15
x=15/9=5/3
no solution: (x-2)and (x-5)>0