Another logarithmic question

**Tutu** · November 23rd 2012, 01:56 AM

Sorry please help me.with another one, how should I do lg(x-2)=(lg3)^2. Thanks really much!

**MarkFL** · November 23rd 2012, 02:17 AM

The notation $\lg$ is not familiar to me (I don't know for sure which base is implied), but I would write:

$\lg(x-2)=\lg(3)\lg(3)$

Now, on the right, use the property $c\cdot\log_a(b)=\log_a(b^c)$ . What do you find?

**Tutu** · November 23rd 2012, 02:34 AM

Thanks sir! lg is log10, has a base 10. but isn't the property you suggested only for one lg? In this case there's two, (lg3)(lg3).

**Prove It** · November 23rd 2012, 04:12 AM

Originally Posted by MarkFL2

The notation $\lg$ is not familiar to me (I don't know for sure which base is implied), but I would write:

$\lg(x-2)=\lg(3)\lg(3)$

Now, on the right, use the property $c\cdot\log_a(b)=\log_a(b^c)$ . What do you find?

You would only be able to use this property if the right hand side was $\displaystyle \begin{align*} \log{(3)} + \log{(3)} = 2\log{(3)} = \log{\left( 3^2 \right)} \end{align*}$ , NOT $\displaystyle \begin{align*} \log{(3)} \log{(3)} = \left[ \log{(3)} \right]^2 \mathbf{\neq} \log{\left( 3^2 \right)} \end{align*}$ .

To solve this equation...

$\displaystyle \begin{align*} \log{(x - 2)} &= \left[ \log{(3)} \right]^2 \\ x - 2 &= 10^{\left[ \log{(3)} \right]^2} \\ x &= 10^{\left[ \log{(3)} \right]^2} + 2 \end{align*}$