Logarithms

[MohtasaUnique]

How do you solve these: log(2x-1) = 2+log(x-2)

 

[Koot]

Subtract the logs to one side and using the log rules, you can combine them.

log A - log B = log A/B

The you can change the logs to 10^ and solve.

 

[panzerjager]

Subtract the logs to one side and using the log rules, you can combine them.

log A - log B = log A/B

The you can change the logs to 10^ and solve.

Yea so : log(2x-1) = 2+log(x-2)
becomes log(2x-1)-log(x-2)= 2
which is: log[(2x-1)/(x-2)] = 2
Turns into: 10^2 = (2x-1)/(x-2) = 100

i dont have a calculator on me but its gonna be a relatively small #

 

[shelnaz2]

dude is it natural log or log base 10?

 

[valvemasher21]

Subtract the log(x-2) to the other side, use the log identity to combine them(log A-log B=log(A/B), and raise 10 to the power of the other side and set it equal to (2x-1)/(x+2).