how do i do this logarithmic problem?

Question

find x such that log(x+50-log(x-1)=2

what are the steps to this problem because it doesn't make sense to m, thanks!

and this problem..which is the same...

log(x+4)-log(x-2)=3

the problem above is
log(x+5)-log(x-1)=2

Answer 1

Do you mean for your parenthesis to be as they are? Because if they are a little bit different the problem becomes much easier.

Answer 2

log(x + 5) - log(x - 1) = 2

First, use the log identity to combine the logs.

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

Presuming this is base 10, convert this to logarithmic form.
Remember that log[base b](a) = c in exponential form is
b^c = a.

10^2 = (x + 5)/(x - 1)
100 = (x + 5)/(x - 1)
100(x - 1) = x + 5
100x - 100 = x + 5
99x = 105
x = 105/99

It's possible that you meant a different base other than base e (because base e has its own name, ln), and if you did, your answer will not be that. I assumed base 10 because log on a calculator is base 10.

Answer 3

log (x+5) - log(x-1) =2
=>
log ((x+5)/(x-1)) = 2

U should be able to go from there though. And since I don't know the log base, I can't continue telling you know to do so.

Answer 4

Subtract means divide logs.

log (x+5 / x-1 )= 2
base is 10

10^2 = x+5/x-1

x-1(100) = x+5

100x - 100 = x+5

99x = 105
x = 105/99

Answer 5

subtracting log's can be substituted into a ratio of logs:

log(x+5)-log(x-1)=2
log(x+5/x-1)=2

Exponentiate both sides:
x-5/x-1=e^2
x-5=xe^2-e^2
x(1-e^2)=e^2-5
x=e^2-5/1-e^2

Plug that into your calculator.