Hi,

Can someone help me to solve this equation for X??

3 Log x - 3 = 2 Log x2(square)

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- Nov 13th 2010, 09:42 AMAniffExponential and Logarithmic Functions
Hi,

Can someone help me to solve this equation for X??

3 Log x - 3 = 2 Log x2(square) - Nov 13th 2010, 09:50 AMearboth
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 $\displaystyle x = 10^{-3}=0.001$