Hi,
Can someone help me to solve this equation for X??
3 Log x - 3 = 2 Log x2(square)
1. Do you mean:
$\displaystyle 3\log(x-3)=2\log(x^2)$
or
$\displaystyle 3\log(x)-3=2\log(x^2)$
2. I'll take the 2nd equation:
$\displaystyle 3\log(x)-3=2\log(x^2)~\implies~3\log(x)-3=4\log(x)$
$\displaystyle -3=\log(x)$
3. Use the sides of the equation as exponents to the base of the logarithms:
If the base is e you'll get $\displaystyle x = e^{-3}\approx 0.0498$
if the base is 10 you'll get $\displaystyle x = 10^{-3}=0.001$