1. ## Logs

SOLVE FOR X:

$\displaystyle \log_{8}{2^3}+\log_{3}\left(x^2-8\right) = \log_{4}\sqrt{16}+\log_{3}\left(2\right)+2\log_{9} {x}$

2. Originally Posted by Sakai
SOLVE FOR X:

$\displaystyle \log_{8}{2^3}+\log_{3}\left(x^2-8\right) = \log_{4}\sqrt{16}+\log_{3}\left(2\right)+2\log_{9} {x}$
Hint: $\displaystyle \log_a a=1$ and $\displaystyle \log_{a^2}x=\tfrac{1}{2}\log_a x$. Do some minor simplifications first and try to get everything in terms of log base 3.

Can you try to proceed?

3. Cool! I didn't know that $\displaystyle \log_{a^2}x=\tfrac{1}{2}\log_a x$.
Originally Posted by Chris L T521
Hint: $\displaystyle \log_a a=1$ and $\displaystyle \log_{a^2}x=\tfrac{1}{2}\log_a x$. Do some minor simplifications first and try to get everything in terms of log base 3.
Can you try to proceed?
So $\displaystyle \log_{8}{2^3} = 1$, $\displaystyle \log_{4}\sqrt{16} = 1$, and $\displaystyle 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}$. I get $\displaystyle \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x}$ which means $\displaystyle 2x = x^2-8$, so we solve the quadratic $\displaystyle x^2-2x-8 = (x-4)(x+2) = 0$ to get $\displaystyle x = 4$ and $\displaystyle x = -2$. Thank you.

4. Originally Posted by Sakai
Cool! I didn't know that $\displaystyle \log_{a^2}x=\tfrac{1}{2}\log_a x$.
Its a consequence of the change of base formula: $\displaystyle \log_{a^2}x=\frac{\ln x}{\ln(a^2)}=\frac{\ln x}{2\ln a}=\frac{1}{2}\left(\frac{\ln x}{\ln a}\right)=\tfrac{1}{2}\log_a x$

So $\displaystyle \log_{8}{2^3} = 1$, $\displaystyle \log_{4}\sqrt{16} = 1$, and $\displaystyle 2\log_{9}{x} = 2\log_{3^2}{x} = \frac{2}{2}\log_{3}{x} = \log_{3}{x}$. I get $\displaystyle \log_{3}\left(x^2-8\right) = \log_{3}{x}+\log_{3}{2} \Leftrightarrow \log_{3}\left(x^2-8\right) = \log_{3}{2x}$ which means $\displaystyle 2x = x^2-8$, so we solve the quadratic $\displaystyle x^2-2x-8 = (x-4)(x+2) = 0$ to get $\displaystyle x = 4$ and $\displaystyle x = -2$. Thank you.
But note that $\displaystyle x=-2$ can't be a solution! Do you see why?

5. Originally Posted by Chris L T521
Its a consequence of the change of base formula: $\displaystyle \log_{a^2}x=\frac{\ln x}{\ln(a^2)}=\frac{\ln x}{2\ln a}=\frac{1}{2}\left(\frac{\ln x}{\ln a}\right)=\tfrac{1}{2}\log_a x$
Thanks. I will need to familiarise myself with that formula.
But note that $\displaystyle x=-2$ can't be a solution! Do you see why?
Because $\displaystyle \log{\left(x\right)}$ is defined only for positive $\displaystyle x$, right?

6. Originally Posted by Sakai
Because $\displaystyle \log{\left(x\right)}$ is defined only for positive $\displaystyle x$, right?
Correct!