# Another logarithmic question

• Nov 23rd 2012, 01:56 AM
Tutu
Another logarithmic question
• Nov 23rd 2012, 02:17 AM
MarkFL
Re: Another logarithmic question
The notation $\displaystyle \lg$ is not familiar to me (I don't know for sure which base is implied), but I would write:

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

Now, on the right, use the property $\displaystyle c\cdot\log_a(b)=\log_a(b^c)$. What do you find?
• Nov 23rd 2012, 02:34 AM
Tutu
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).
• Nov 23rd 2012, 04:12 AM
Prove It
Re: Another logarithmic question
Quote:

Originally Posted by MarkFL2
The notation $\displaystyle \lg$ is not familiar to me (I don't know for sure which base is implied), but I would write:

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

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

You would only be able to use this property if the right hand side was \displaystyle \displaystyle \begin{align*} \log{(3)} + \log{(3)} = 2\log{(3)} = \log{\left( 3^2 \right)} \end{align*}, NOT \displaystyle \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 \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*}
• Nov 23rd 2012, 10:07 AM
MarkFL
Re: Another logarithmic question
I was suggesting to write:

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

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

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