Exponential and Logarithmic Functions

**Aniff** · November 13th 2010, 09:42 AM

Hi,

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

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

**earboth** · November 13th 2010, 09:50 AM

Originally Posted by Aniff

Hi,

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

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

1. Do you mean:

$3\log(x-3)=2\log(x^2)$
or
$3\log(x)-3=2\log(x^2)$

2. I'll take the 2nd equation:

$3\log(x)-3=2\log(x^2)~\implies~3\log(x)-3=4\log(x)$

$-3=\log(x)$

3. Use the sides of the equation as exponents to the base of the logarithms:

If the base is e you'll get $x = e^{-3}\approx 0.0498$

if the base is 10 you'll get $x = 10^{-3}=0.001$

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