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

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- Nov 23rd 2012, 01:56 AMTutuAnother logarithmic question
Sorry please help me.with another one, how should I do lg(x-2)=(lg3)^2. Thanks really much!

- Nov 23rd 2012, 02:17 AMMarkFLRe: 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 AMTutuRe: 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 AMProve ItRe: Another logarithmic question
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 AMMarkFLRe: Another logarithmic question
I was suggesting to write:

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

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

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