# 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) =...

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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.

Approved by eNotes Editorial Team 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

Approved by eNotes Editorial Team