Does anyone know how to solve log (x-5) =2?

Answer 1

use the properties of logarithm

I'll assume that it is in base 10

10^y = a property 1

log a = y property 2

in your problem log (x-5) =2

let a = x-5 and y =2

just substitute the variables of property 1 namely the y and a

10^ 2 = x -5

100 = x-5

x=105

Answer 2

Assuming the logarithm used is to the base e, where e is exp 1,
taking the exponential of both sides gives
e^[log (x-5)] = e^2
=> x - 5 = e^2
=> x = 5 + e^2

Answer 3

If it's logs to the base e, then...

log (x-5) = 2

(x - 5) = e^2

x = e^2 + 5

If it's a different base, then use that instead of e.

Answer 4

x-5 = e^ 2 using ( log x = m, then x = Antilog m or x = e^ m)
x= 5 +2.7^2 = 12.29