Solve algebraically for x: log x^2 + log x^-1 = 2
You may use other property of logarithms: the addition of two logarithms that have the same bases becomes a product.
property: log a + log b = log (a*b)
log x^2 + log x^-1 = log (x^2*x^-1)
use the property of negative powers:
x^-1 = 1/x => log (x^2*x^-1) = log (x^2/x) = log x
Equate log x = 2 and take antilog to find out x.
x = 10^2 => x = 100
Answer: Solution of logarithmic equation is x=100.
We have to solve : log x^2 + log x^-1 = 2
log x^2 + log x^-1 = 2
use the property of logarithm: log x^a = a*log x
=> 2*log x - log x = 2
=> log x = 2
=> x = 10^2
=> x = 100
The required value of x = 100