2. ## Re: Another logarithmic question

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?

3. ## Re: Another logarithmic question

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).

4. ## Re: Another logarithmic question

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*}

5. ## Re: Another logarithmic question

I was suggesting to write:

$\lg(x-2)=\lg\left(3^{\lg(3)} \right)$

$x-2=3^{\lg(3)}$

$x=2+3^{\lg(3)}$