1. ## understanding logs

Hello,

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.

2. Originally Posted by l flipboi l
Hello,

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.
Notice that $\displaystyle 8^{\frac{1}{2}} = 4$.

Therefore $\displaystyle 8^{-\frac{1}{2}} = \frac{1}{4}$.

What do you think $\displaystyle x$ equals?

3. not sure =(

4. Look at the question carefully.

I have told you that $\displaystyle 8^{-\frac{1}{2}} = \frac{1}{4}$.

You are told that $\displaystyle 8^x = \frac{1}{4}$.

Therefore $\displaystyle 8^x = 8^{-\frac{1}{2}}$.

What do you think $\displaystyle x$ equals?

5. Prove It : $\displaystyle 8^{\frac{1}{2}} = \sqrt{8} \neq 4$

The $\displaystyle x$ is already isolated, so you just have to evaluate. Do you need help about evaluating it ?

6. the answer in the book is 2/3. I'm trying to figure out how to get this. so i looked up raising a power fractions and should it be

sqrt(8) ?

7. yes, i need to see how to evaluate it.

8. 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)

9. great, this is what i was looking for. Thanks!

10. Originally Posted by Bacterius
Prove It : $\displaystyle 8^{\frac{1}{2}} = \sqrt{8} \neq 4$

The $\displaystyle x$ is already isolated, so you just have to evaluate. Do you need help about evaluating it ?
Hahaha oops. Of course it doesn't

Sorry, massive brain fart there...

11. Originally Posted by l flipboi l
Hello,

log8(1/4) = x

8^x = 1/4

how would i solve this using algebra?
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.

This is simply review and i never really learned how to do it.
Learn the definitions first!

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.

12. Originally Posted by Prove It
Notice that $\displaystyle 8^{\frac{1}{2}} = 4$.

Therefore $\displaystyle 8^{-\frac{1}{2}} = \frac{1}{4}$.

What do you think $\displaystyle x$ equals?
What? That's not true at all. $\displaystyle 8^{1/2}= \sqrt{8}= 2\sqrt{2}$, not 4! 8 times 1/2= 4.

13. Originally Posted by HallsofIvy
What? That's not true at all. $\displaystyle 8^{1/2}= \sqrt{8}= 2\sqrt{2}$, not 4! 8 times 1/2= 4.

14. ## understanding logs

posted by flipboi

log base8(1/4)=x find x
8^x=1/4
xlogbase2(8) =log base2(1/4)
xtimes3=-2
x=-2/3
Easier.All logs bear a fixed relationship for all bases.You can pick your own to suit a problem

bjh

15. Originally Posted by Prove It