Hello,
Can you please help me understand how to solve this...
log8(1/4) = x
8^x = 1/4
how would i solve this using algebra?
This is simply review and i never really learned how to do it.
The change of base formula allows us to compute any base logarithms from another one. That is :
$\displaystyle \log_a{x} = \frac{\log_b{x}}{\log_b{a}}$
So you plug in your values : you have $\displaystyle a = 8$, and you want to be able to calculate this with your calculator. Your calculator (hopefully) has a "log" button, so we'll use this one. It's the decimal logarithm, so $\displaystyle b = 10$. And we want to compute $\displaystyle x = \frac{1}{4}$. So :
$\displaystyle \log_8{\frac{1}{4}} = \frac{\log_{10}{\frac{1}{4}}}{\log_{10}{8}}$
And this gives exactly $\displaystyle \log_8{\frac{1}{4}} = - \frac{2}{3}$
This can be verified easily : it is equivalent to saying that $\displaystyle \frac{1}{4} = 8^{- \frac{2}{3}}$, which is true (you can check)
You wouldn't. You would use the definition of "logarithm to base a" which is that it is the "inverse function" to $\displaystyle a^x$. The logarithm is defined by "if $\displaystyle y= log_a(x)$ then $\displaystyle x= a^y$".
That's all you need.
Learn the definitions first!This is simply review and i never really learned how to do it.
Prove it, in addition to simply giving a wrong calculation, seems to think you were trying to solve $\displaystyle 8^x= 1/4$ but the solution to that is $\displaystyle x= log_8 (1/4)$ which is your first equation. I assume you are really asking how to go from one equation to the other.
If you are really asking how to write $\displaystyle x= log_8(1/4)$ in terms your calculator can handle,use the fact that is $\displaystyle 8^x= 1/4$ then $\displaystyle log(8^x)= xlog(8)= log(1/4)$ so that $\displaystyle x= \frac{log(1/4)}{log(8)}$ where "log" can be either base 10 (common log) or base e (natural log), both of which should be on your calculator.