Asked by Anonymous
                find the values of x which satisfy the equation log_10(x^2+4) = 2+log_10 x-log_10 20
            
            
        Answers
                    Answered by
            mathhelper
            
    If the base of your log is 10, we don't have to write the base, so ...
log(x^2+4) = 2 + log x - log 20
log(x^2 + 4) - log x + log 20 = 2
log( 20(x^2 + 4)/x ) = log 100
20(x^2 + 4)/x = 100
(x^2 + 4)/x = 5
x^2 + 4 = 5x
x^2 - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4
    
log(x^2+4) = 2 + log x - log 20
log(x^2 + 4) - log x + log 20 = 2
log( 20(x^2 + 4)/x ) = log 100
20(x^2 + 4)/x = 100
(x^2 + 4)/x = 5
x^2 + 4 = 5x
x^2 - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4
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