Thread: log function....

how would i get x? the previous method does not work.....

You already have x!

Originally Posted by pickslides

You already have x!

well i meant in its simplest form...

It doesn't get much more simple

$\displaystyle x = 2^{\log_23+\log_28}$

$\displaystyle x = 2^{\log_23+\log_22^3}$

$\displaystyle x = 2^{\log_23+3\log_22}$

$\displaystyle x = 2^{\log_23+3\times 1}$

$\displaystyle x = 2^{\log_23+3}$

$\displaystyle x = 2^3\times 2^{\log_23}$

$\displaystyle x = 8\times 2^{\log_23}$

i'm supposed to be getting a whole number i think

Well it's nearly there. What's $\displaystyle 2^{\log_2 x}$ for any $\displaystyle x$?

Originally Posted by Matt Westwood

Well it's nearly there. What's $\displaystyle 2^{\log_2 x}$ for any $\displaystyle x$?

honestly i'm not too sure, my guess would be 1?

Hello, frozenflames!

Simplify: .$\displaystyle x \:=\:2^{\log_23 + \log_28} $

We have: .$\displaystyle x \;=\;2^{\log_23+\log_28} \;=\; 2^{\log_2(3\cdot8)} \;=\;2^{\log_2(24)} \;=\;24 $

Originally Posted by frozenflames

honestly i'm not too sure, my guess would be 1?

This is the most fundamental truth about exponentials and logarithms that you really need to understand: one is the inverse of the other. That is, logarithms "undo" exponents, and exponents "undo" logarithms.

$\displaystyle a = \log_b c \iff c = b^a$ for any $\displaystyle a$, and any $\displaystyle b, c$ positive. That's how logarithms are defined.

So $\displaystyle 2^{\log_2 x} = \log_2 (2^x) = x$.